subject:"Re\: Why are laws of physics stable\?"


> On 10 Jul 2021, at 21:48, 'Brent Meeker' via Everything List 
>  wrote:
> 
> 
> 
> On 7/10/2021 2:00 AM, Bruno Marchal wrote:
>> 
>>> On 6 Jul 2021, at 22:07, 'Brent Meeker' via Everything List 
>>> >> > wrote:
>>> 
>>> 
>>> On 7/6/2021 10:34 AM, Jason Resch wrote:
 
 
 On Tue, Jul 6, 2021 at 12:27 PM 'Brent Meeker' via Everything List 
 >>> > wrote:
 And you're never going to find a being that behaves intelligently based on 
 information that can be quantum erased.
 
 You need only a quantum computer with enough qubits.
>>> Can you prove that?  How does this quantum intelligence ever arrive at a 
>>> definite decision?
>>> 
>>> 
>> 
>> With a quantum brain, you can hack all credit cards, by running shor 
>> algorithm in your head for example.
>> 
>> You argument seems to negate the possibility of quantum speeding. You would 
>> be right if P = NP, or something…
> 
> No.  As I recall even Shor's algorithm has a probabilistic step so that the 
> answer is only correct with high probability, not certainty. 

Right.


> Quantum speedup is possible, just not quantum definiteness.  One you or your 
> QC decides to act that act cannot be a superposition of actions. 

Why? It cannot be a superposition of personal experience, but if I decide in 
Helsinki to go to W or to M with a quantum coin, then, without wave collapse, 
my actions (going to W) and (going to M) will be superposed, despite each of me 
will not been able to fell it, for obvious mechanical reason.



> And the decoherence is not just FAPP, it is  inherent in the loss of 
> information across the Hubble boundary.

But the informations does not disappear. It becomes relatively inaccessible, 
only.

Bruno





> 
> Brent
> 
>> 
>> Bruno
>> 
>> 
>>> Brent
>>> 
>>> 
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Re: Why are laws of physics stable?


> On 10 Jul 2021, at 21:45, 'Brent Meeker' via Everything List 
>  wrote:
> 
> 
> 
> On 7/10/2021 1:48 AM, Bruno Marchal wrote:
>> 
>>> On 6 Jul 2021, at 16:45, Tomas Pales >> > wrote:
>>> 
>>> 
>>> On Tuesday, July 6, 2021 at 10:28:11 AM UTC+2 Bruno Marchal wrote:
>>> 
 On 3 Jul 2021, at 14:13, Tomas Pales >>> > wrote:
 Can't there be a machine that computes gravitational interaction with 
 gravitational constant 6.674 x 10 to the -11 up until some time t and then 
 continues the computation with gravitational constant 5 x 10 to the -11, 
 or just halts? That would be an instability or cessation of gravitational 
 law.
>>> 
>>> Yes, and that exists, but such world will have a very low probability to be 
>>> accessed by any observer, due to the fact that, below our mechanist 
>>> substitution level, all such theories intervene.
>>> 
>>> How low is this probability? Is it maybe as low as the probability that my 
>>> whole body quantum-tunnels through a wall?
>> 
>> Yes.
>> 
>> 
>> 
>>> 
>>> What does it mean that "below our mechanist substitution level, all such 
>>> theories intervene”?
>> 
>> 
>> If your consciousness does not require some details: like the position of an 
>> electron of some atom in some neurotransmitter (say), then it will be 
>> associate as much with your brain and that election here, and with you brain 
>> and that electron there, and if you don’t measure the position of that 
>> electron, the two histories can interfere statistically.
> What would it mean to measure the state of your own brain?

To encode the relevant information such that I survive the copy/reconstitution. 
The possibility of this is assumed in the YD part of the YD+CT.
YD = “Yes doctor”, and CT = Church’s Thesis (or Church-Turing’s thesis).

Bruno




> 
> Brent
> 
>> 
>> An electronic orbital is such a map: it tells you where you can find the 
>> electron in your most probable computational histories.
>> 
>> Bruno 
>> 
>> 
>> 
>> 
>>> 
>>> 
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Re: Why are laws of physics stable?


> On 10 Jul 2021, at 21:43, 'Brent Meeker' via Everything List 
>  wrote:
> 
> 
> 
> On 7/10/2021 1:28 AM, Bruno Marchal wrote:
>> 
>>> On 6 Jul 2021, at 12:55, John Clark >> > wrote:
>>> 
>>> 
>>> On Sun, Jul 4, 2021 at 10:10 PM 'Brent Meeker' via Everything List 
>>> >> > wrote:
>>> 
>>> >> It's easy to determine that the quantum computer is intelligent but as 
>>> >> for consciousness, how did you determine that it was not conscious? For 
>>> >> that matter how did you determine that I am conscious? But let's get out 
>>> >> of the consciousness quagmire for a moment so I can ask you a question, 
>>> >> leaving behind the interpretation of the experiment concentrating only 
>>> >> on its results, if it was actually performed as described do you think 
>>> >> interference bands would be on that photographic plate or would there be 
>>> >> no such bands? I would bet money the bands would be there on that plate 
>>> >> even though there's no longer any which way information remaining. So, 
>>> >> what would you put your money on, bands or no bands? 
>>> > I would guess the interference bands would be present exactly because, ex 
>>> > hypothesi, the which-way information was quantum erased.
>>> 
>>> So an intelligent and presumably conscious being once existed that knew 
>>> which slot all the electrons went through, but those interference bands 
>>> still showed up anyway. Don't you find that a little strange? If Many 
>>> Worlds is wrong and that being didn't exist in another world, then where 
>>> did it exist? 

I didn't wrote that.

Bruno


> It, or rather it's knowledge, existed in the past, before being erased.
> 
> Brent
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Re: Why are laws of physics stable?



> On 10 Jul 2021, at 21:41, 'Brent Meeker' via Everything List 
>  wrote:
> 
> 
> 
> On 7/10/2021 1:19 AM, Bruno Marchal wrote:
>> So, in general we can conclude by generalizing this to any large number of 
>> particles that even with what we consider to be permanent records, you don't 
>> get rid of the theoretical possibility of interference between the sectors 
>> where those records are different.

I did not write this, and out of context, I don’t know if I agree or not with 
this. 


> 
> We can if the universe is expanding faster than light beyond the Hubble 
> radius.

How do you a physical being to singularise its consciousness? 

The term “physical universe” is no more an ontological being if we assume 
Digital Mechanism. It is an appearance, a phenomenological reality, not an 
ontology.

Bruno



> 
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Re: Why are laws of physics stable?


> On 10 Jul 2021, at 21:38, 'Brent Meeker' via Everything List 
>  wrote:
> 
> 
> 
> On 7/10/2021 1:06 AM, Bruno Marchal wrote:
>> 
>>> On 5 Jul 2021, at 21:01, 'Brent Meeker' via Everything List 
>>> >> > wrote:
>>> 
>>> 
>>> On 7/5/2021 7:41 AM, John Clark wrote:
 On Mon, Jul 5, 2021 at 9:44 AM Tomas Pales >>> > wrote:
 
  >> "Brain" is a noun, "consciousness" is not, that's why you can't 
 measure consciousness by the pound or by the cubic inch.
 
 > In English language it is used as a noun. Check out a dictionary:
 
 consciousness noun 
 I know, that's what my fourth grade teacher told me too, but I long-ago 
 realized that neither she nor the lexicographers who wrote that big thick 
 book are the fonts of all wisdom.
 >> Intelligence is what a brain does not what a brain is, and because 
 >> Darwinian Evolution is almost certainly correct, consciousness must be 
 >> an inevitable byproduct of intelligence, therefore "consciousness" is 
 >> not a noun, it's a word that describes what a noun (in this case the 
 >> brain) does, in other words consciousness is an adjective.
 
 > You mean a verb then, no?
 
 I think adjective fits the bill a little better, I think Tomas Pales is 
 the way atoms behave when they are arranged in a Tomaspalesian way.
 
  > consciousness is a spatiotemporal object.
 
 I disagree, I think asking where my consciousness is located would be like 
 asking where the number 11 or the color yellow  or "fast" is located.  If 
 my brain is in Paris and I'm looking at a TV football game from Detroit 
 and I'm listening to a friend in Australia on my telephone and I'm 
 thinking about The Great Wall of China would it  make sense to say my 
 consciousness is really located inside a box made of bone mounted on my 
 shoulders when I have no conscious experience of being in a bone box on my 
 shoulders? I don't think so.
>>> Yet a sharp blow to that bone box would eliminate your conscious experience 
>>> at least temporarily. 
>>> 
>> Only from the point of view of some conscious subject. From the point of 
>> view of the person associated to the brain in the box, that does not make 
>> sense, as it is associated to infinitely many truing universal relation.
> 
> That's incorrect.  I've been knocked unconscious and when I regained 
> consciousness (it was on a few seconds) I realized the gap my conscious 
> experience.

That does not entail that there were a gap. That entails only that at some 
moment you experience a feeling that there was a gap, from which you infer that 
there was a gap, but maybe you are just amnesic about your consciousness during 
the gap, or perhaps, you were really unconscious, but by definition, that is 
not part of the experience.
Anyway, my point is that you survived the knocking. Not that you feel there has 
been a gap, which by the way, confirms your first person survival.







> 
>> 
>> The body is only a map on infinitely many histories. That can be proved both 
>> with QM-without-collapse, or in any non trivial combinatory algebra (like a 
>> model of arithmetic).
>> 
>> 
>>> So there's something there that is essential to your consciousness.
>> 
>> What is “essential” are the infinitely many computations.
>> 
>> Since the 1930s we know that all computations are realised in any model (in 
>> the logician sense) of arithmetic, or of combinatory logic (Kxy = x, Sxyz = 
>> xz(yz)).
> 
> But not in any brain...they are only finite.

The theories, words, axioms, machines, brains are all finite, but the semantic 
are not, the model of any Turing complete theory is always infinite, and 
provably realise all finite and infinite computation.
String and Gravity loop theories are also finite, independently that their 
model are all infinite.



> 
>> 
>> I know that this contradict 1492 years of materialist brainwashing, but 
>> “appearance of matter” are explained in arithmetic, and get contradictory 
>> when associated or singularised through any supplementary axioms, even the 
>> induction axioms used to define what an observer can be.
>> 
>> You assume some ontological commitment inconsistent with Mechanism here.
> 
> You assume an ontological commitment to Church-Turing infinite computations.  

No. I assume Elementary Arithmetic, i.e.the natural numbers, or the combinator, 
and their basic laws, or any Turing universal machinery (the phi_i). 
All physical theories assumes more than this. By assuming less, we lost the 
ability token define what is a digital machine.

The existence of the infinite computations is then proved  by the Löbian 
machine simulated by the numbers (sigma_1) relations, who have the induction 
axiom, and can handle the infinite, like PA can prove the existence of an 
infinity of prime numbers, or

Re: Why are laws of physics stable?

> On 10 Jul 2021, at 14:17, John Clark  wrote:
> 
> On Sat, Jul 10, 2021 at 3:52 AM Bruno Marchal  > wrote:
> 
> >> "Brain" is a noun, "consciousness" is not
>  
> > I disagree with this
> 
> An embalmed brain rotting in a grave is a noun. Do you therefore think it's 
> conscious?

A brain is not conscious. Only a person is conscious, and the brain has some 
role here. Anyways this does not make the term “consciousness" not being a noun.

> I don't because it's not doing anything that 3 pounds of rotting hamburger 
> isn't doing, and neither of the two are behaving intelligently. 
> 
> >> therefore "consciousness" is not a noun, it's a word that describes what a 
> >> noun (in this case the brain) does, in other words consciousness is an 
> >> adject
> 
> > This is logically inconsistent with Descartes Mechanism. 
> 
> I have no idea what  "Descartes Mechanism" is, and after listening to you all 
> these years I am quite certain you can't give a coherent explanation for it 
> either, but whatever it means if it is logically inconsistent with what I 
> said then "Descartes Mechanism" is wrong.

Descartes’s Mechanism is the idea that the human (and animal’s) body is a 
(natural) machine. By (indexical, digital) mechanism, aka computationalism,  I 
mean since the beginning the thesis that we can survive with a digital brain (a 
physical computer) transplantation, like most people believe that we can 
survive with an artificial heart.

Then the consequence is that the min body problem is reduced to a statistics on 
the infinitely many computations going through our actual states. This makes 
the logic of the observable having to obey to the modes []p & <>t & p, and some 
others, and that works as we get both the many-histories view on the physical 
reality, and its quantum formalism. 
We cannot prove mechanism, but we can count the evidence for.

> > Without mechanism, it is consistent, but still problematic with Occam razor 
> 
> As I said before,  Occam razor is about economy of assumptions not economy of 
> results. 

Absolutely. In this case Mechanism win, because the theory of everything can be 
just the two equations/assumptions:

Kxy = x
Sxyz = xz(yz)

+ identity rules (as I have described in detail last year)

Everett assumes this implicitly, but assume also the Wave Equation, when 
actually Mechanism enforces that it has to be derive from the “many-worlds” 
interpretation of elementary arithmetic, or combinator (like K and S), etc.

In that theory, using Mechanism, we can prove the existence of all universal 
machine, and of all the computations, and we extract the theology of the 
Gödel-Löbian machine, whose main axioms is Löb’s formula ([]([]p -> p) -> []p).

My point is that we cannot use an ontological commitment to prevent the logical 
consequence of mechanism, once we assume it..

Bruno

> 
> John K ClarkSee what's on my new list at  Extropolis 
> 
> 0o6
> 
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Re: Why are laws of physics stable?

On Mon, Jul 12, 2021 at 12:44 PM smitra  wrote:

> On 12-07-2021 00:43, Bruce Kellett wrote:
> >
> > You can't decide to put mirrors in place after the photons have
> been emitted.
> > Of course, you could always decide in advance to perform all
> > experiments in reflecting boxes, but that is not what is done in
> > practice. So the general rule is that escaping photons are not
> > recoverable.
>
> If in one experiment with mirrors they are recoverable and therefore
> interference can be detected in principle, then that proves that the
> other sectors do objectively exist. Since the objective existence of
> these sectors does not depend on whether or not one uses mirrors in the
> experiment, one can conclude that they exist in general.
>

The point of measurement is to get a definite permanent result. Escaping
photons (IR or other) are secondary to this endeavour. They just represent
one universal way in which information can escape from an experiment and
contribute to the overall decoherence. If you recover all these photons,
and reverse all other records of the particular experimental result, then
you have effectively not made a measurement. Measurement being defined by
the formation of permanent records of a result. If there are no such
records, then no measurement has been made. That may well be the case. But
that does not demonstrate the objective existence of other sectors, or
other worlds in which different results were obtained. There was no
measurement in that case, so no results were obtained anywhere. Showing
that some particular experiment is (or was) reversible, does not actually
prove anything other than that the laws of physics are reversible. Which we
knew anyway, and is not relevant for measurement.

Bruce

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Re: Why are laws of physics stable?

On Mon, Jul 12, 2021 at 1:48 PM 'Brent Meeker' via Everything List <
everything-list@googlegroups.com> wrote:

>
> So, then you agree with Bruce that so long as the which-way
> information
> exists in those outgoing photons, the interference pattern will show
> up?
>
>
> I think you might have meant that as long as the which-way information
> exists (has not been quantum erased), then no interference is visible.
>
>
> Right.  That's what I meant to write.
>
> In quantum erasure double-slit experiments the which-way photons must be
> focused to the same point regardless of which slit they indicate...that's
> the "erasure".
>

Maybe that is what Saibal is getting at with his observation of the photons
at some point on a second screen. His notation is for a mechanism at each
slit which emits a photon if the ball went through that slit. If the
photons emitted from either slit are observed at the same point on the
photon screen, then you can't tell from which slit they originated -- hence
the "erasure". That makes sense, but it did not seem to be what Saibal was
saying. And it bears little relation to the actual setup in the buckyball
experiment. There is no mechanism to register which slit the ball went
through -- the photn(s), if any, are emitted downstream as the heated ball
transitions to a lower energy state.

Bruce

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Re: Why are laws of physics stable?

On Mon, Jul 12, 2021 at 1:48 PM 'Brent Meeker' via Everything List <
everything-list@googlegroups.com> wrote:

> On 7/11/2021 7:57 PM, smitra wrote:
>
>
> It's a correct statement, see https://doi.org/10.1071/AS03040 page 101
> section 3.3:
>
> "The most distant objects that we can see
> now were outside the Hubble sphere when their comoving
> coordinates intersected our past light cone. Thus, they
> were receding superluminally when they emitted the photons
> we see now. Since their worldlines have always been
> beyond the Hubble sphere these objects were, are, and
> always have been, receding from us faster than the speed
> of light.3"
>
>
> Interesting.  I shall read it carefully.  But it doesn't affect the point
> that in the Bucky Ball double slit experiment the photons can either hit
> the walls or escape to infinity and in either case the interference pattern
> will be washed out.  Even your citation says,
>
> *"No observer ever overtakes a light beam and all observers measure light
> locally to be travelling at c." *Brent
>

The situation is confusing because the Hubble radius expands with time, so
objects (and light) that were once beyond the Hubble horizon can enter it
and be seen now. I think the important point is "The particle horizon can
be larger than the event horizon, although we cannot see events that occur
beyond our current event horizon, we can still see many galaxies that are
beyond our current event horizon by light they emitted long ago."

This does not alter the fact that light that we emit now and that passes
out beyond our Hubble horizon is lost from us forever. And events that now
occur beyond our Hubble horizon are forever hidden from us. In other words,
the photons escaping in the buckyball experiment are lost beyond
possibility of recovery. So the which-way information they carry still
exists somewhere, and the interference is washed out, as Brent says.

Bruce

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Re: Why are laws of physics stable?

2021-07-11 Thread 'Brent Meeker' via Everything List

On 7/11/2021 7:57 PM, smitra wrote:

On 12-07-2021 00:35, Bruce Kellett wrote:

On Mon, Jul 12, 2021 at 7:13 AM 'Brent Meeker' via Everything List
wrote:

On 7/11/2021 9:53 AM, smitra wrote:

On 11-07-2021 02:41, 'Brent Meeker' via Everything List wrote:

On 7/10/2021 5:35 PM, smitra wrote:

On 11-07-2021 01:05, Bruce Kellett wrote:

On Sun, Jul 11, 2021 at 1:18 AM smitra

wrote:

On 10-07-2021 07:58, Bruce Kellett wrote:

Neither of your analyses actually explain the observed

behaviour.

It does so, otherwise the experiment would have proven that QM

invalid.

So, in your opinion, the quantum mechanical analysis given in

the

paper is wrong.

It is correct, it's just that your conclusion about the MWI

based on

that paper is incorrect. Escaping infrared photons and the
decoherence this causes are totally irrelevant. Whether or not

interference pattern can be detected is not relevant to the

question

of whether or not a superposition exists when we know that it

exist

and it has no decohered. It's just that you then can't reproduce

one

particular line of evidence for the validity of quantum

mechanics in

that particular experiment.

If I prepare the state of a particle in a superposition and let

this interact in a certain way, then we know how this state will

evolve. If this evolution involves interactions with many

particles,

then the system will decohere. Then it may be true that we

cannot

distinguish the state of the system from being in a pure or

mixed

state in practice due to not being able to conduct an

interference

experiment involving a very large number of particles, but

quantum

mechanics still tells us that the superposition exists and that

we were to conduct the right sort of interference experiment, we

would see an interference that would prove that the state is not

mixed state.

But does it exist if part of the information has crossed the

Hubble

boundary? Then there is no experiment, even in principle, that

would

prove the state is not a mixed one.

The theories that tell you that there is such a thing like a

Hubble

boundary, also tell you that information does not vanish. We can

now

detect photons from galaxies that have always receded from us

faster

than light.

?? I think that's a contradiction.

It's worse than that -- it's gibberish.

It's a correct statement, see https://doi.org/10.1071/AS03040 page 101
section 3.3:

"The most distant objects that we can see
now were outside the Hubble sphere when their comoving
coordinates intersected our past light cone. Thus, they
were receding superluminally when they emitted the photons
we see now. Since their worldlines have always been
beyond the Hubble sphere these objects were, are, and
always have been, receding from us faster than the speed
of light.3"

Interesting. I shall read it carefully. But it doesn't affect the
point that in the Bucky Ball double slit experiment the photons can
either hit the walls or escape to infinity and in either case the
interference pattern will be washed out. Even your citation says, /"No
observer ever overtakes a light beam and all observers measure light
locally to be travelling at c."

/Brent/
/

Saibal

So, that's information we're that has leaked away from those

galaxies

that does objectively exists. To prove that a state is a pure

state

one can just point to the way the state was prepared and invoke
quantum mechanics. It's not necessary to demonstrate interference

prove that quantum mechanics is still valid.

So, then you agree with Bruce that so long as the which-way
information
exists in those outgoing photons, the interference pattern will show
up?

I think you might have meant that as long as the which-way information
exists (has not been quantum erased), then no interference is visible.

Right. That's what I meant to write.

In quantum erasure double-slit experiments the which-way photons must be
focused to the same point regardless of which slit they
indicate...that's the "erasure".

Brent

Bruce

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Re: Why are laws of physics stable?

On 12-07-2021 00:35, Bruce Kellett wrote:

On Mon, Jul 12, 2021 at 7:13 AM 'Brent Meeker' via Everything List
wrote:

On 7/11/2021 9:53 AM, smitra wrote:

On 11-07-2021 02:41, 'Brent Meeker' via Everything List wrote:

On 7/10/2021 5:35 PM, smitra wrote:

On 11-07-2021 01:05, Bruce Kellett wrote:

On Sun, Jul 11, 2021 at 1:18 AM smitra

wrote:

On 10-07-2021 07:58, Bruce Kellett wrote:

Neither of your analyses actually explain the observed

behaviour.

It does so, otherwise the experiment would have proven that QM

invalid.

So, in your opinion, the quantum mechanical analysis given in

the

paper is wrong.

It is correct, it's just that your conclusion about the MWI

based on

that paper is incorrect. Escaping infrared photons and the
decoherence this causes are totally irrelevant. Whether or not

interference pattern can be detected is not relevant to the

question

of whether or not a superposition exists when we know that it

exist

and it has no decohered. It's just that you then can't reproduce

one

particular line of evidence for the validity of quantum

mechanics in

that particular experiment.

If I prepare the state of a particle in a superposition and let

this interact in a certain way, then we know how this state will

evolve. If this evolution involves interactions with many

particles,

then the system will decohere. Then it may be true that we

cannot

distinguish the state of the system from being in a pure or

mixed

state in practice due to not being able to conduct an

interference

experiment involving a very large number of particles, but

quantum

mechanics still tells us that the superposition exists and that

we were to conduct the right sort of interference experiment, we

would see an interference that would prove that the state is not

mixed state.

But does it exist if part of the information has crossed the

Hubble

boundary? Then there is no experiment, even in principle, that

would

prove the state is not a mixed one.

The theories that tell you that there is such a thing like a

Hubble

boundary, also tell you that information does not vanish. We can

now

detect photons from galaxies that have always receded from us

faster

than light.

?? I think that's a contradiction.

It's worse than that -- it's gibberish.

It's a correct statement, see https://doi.org/10.1071/AS03040 page 101
section 3.3:

Saibal

So, that's information we're that has leaked away from those

galaxies

that does objectively exists. To prove that a state is a pure

state

one can just point to the way the state was prepared and invoke
quantum mechanics. It's not necessary to demonstrate interference

prove that quantum mechanics is still valid.

So, then you agree with Bruce that so long as the which-way
information
exists in those outgoing photons, the interference pattern will show
up?

I think you might have meant that as long as the which-way information
exists (has not been quantum erased), then no interference is visible.

Bruce

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Re: Why are laws of physics stable?


On 12-07-2021 03:42, Bruce Kellett wrote:

On Mon, Jul 12, 2021 at 8:43 AM Bruce Kellett 
wrote:


On Mon, Jul 12, 2021 at 3:44 AM smitra  wrote:


On 11-07-2021 02:46, Bruce Kellett wrote:




This is where you go seriously wrong. Simply recording where the
photons land does not quantum erase the information they

carried. Once

the photons carry off the which way information, the

interference

pattern is restored only if the information carried by the

photons is

quantum erased. Simply running the photons into a screen (or the
wall), even if you record where they land, is not quantum

erasure.

See, for example, the paper arXiv:1206.6578 on quantum erasure.

In

this paper they say "the presence of path information anywhere

in the

universe is sufficient to prohibit any possibility of

interference. In

other words, the atoms' path states alone are not in a coherent
superposition due to the atom-photon entanglement." This

transfers

directly to the buckyball experiment under discussion. Running

the

photons into a screen, or the wall, does not destroy the

ball-photon

entanglement.



Recording where the photons land on the screen is enough, this is
very
easy to see. Let's consider an interference experiment where the
particle gets entangled with another particle that carries away
the
which way information. If we work in the position representation
then we
have a wavefunction:

psi(x,y) = 1/sqrt(2) [psi_1(x,y) + psi_2(x,y)](1)

where x is the position of particle 1 just before it hits the
screen and
y the position of the other particle at that time, and psi_1(x,y)
is the
wavefunction when only slit 1 is open while psi_2(x,y) the
wavefunction
with only slit 2 open. We also assume that particle 2 carries the
which
way information, this means that psi_1(x,y) and psi_2(x,y) for x
kept
fixed and considered as a function of y are eigenfunctions with
different eigenvalues of an observable that corresponds to
extracting
the which way information from the second particle. This then
implies
that psi_1(x,y) and psi_2(x,y) are orthogonal for every x:

Integral psi_1(x,y)*psi_2(x,y)d^3y = 0 (2)


This is the same mistake that you have routinely made in all of
these discussions. You calculate for particles through each slit
independently. But if that is the case, then no interference can
ever be seen. The possibility of interference relies on coherence
between the amplitudes at the two slits. In your calculations, you
have never allowed for this coherence -- the amplitudes at the two
slits are not independent.


To see this in more detail, we can go through your original argument.
You write : denote the buckyball moving through the left slit by |L>
and the right slit by |R> and the photons by |PL> and |PR>. Now the
states |R> and |L> are coherent; thus correlated; and the overlap
 does not vanish. Writing

 |psi> = (|L>|PL>  +  |R>|PR>)

is the same as assuming that the photons are emitted from the slits if
they detect a ball passing through that slit. This is a direct
"which-way" measurement, and since the photon states are independent,
they are, as you say, orthogonal, and there can be no interference if
such measurements are made at the slits. This is the essential loss of
coherence.

Your claim then is that if we observe the photons on another screen
and keep track of the ball positions on their screen for photons that
arrive at the same point on this other screen, then the interference
pattern is restored. This is wrong, because the photons emitted from
the slits if and when they detect a ball passing through the
corresponding slit are still independent. The fact that they might
arrive at the same point on a further screen is irrelevant. The photon
states are still independent. In fact, there can only ever be one
photon for each buckyball, so there can be no case in which two
photons arrive at the photon screen at the same time. So the photon
states are necessarily always independent and orthogonal, and your
overlap function

   = 0,  always.



No, we perform a detection at a point y on the photon screen, the 
product of these amplitudes are not zero, it only becomes zero if you 
integrate this over y. The rest of your argument does not apply


Saibal


Consequently, your joint detection of photons does not restore the
interference. Decoherence, once present, destroys coherence. And this
coherence cannot be restored simply by observing the which-way
photons. You have to quantum erase the which-way information. And this
is not easy for this particular set-up. Quantum erasure essentially
requires a measurement of a conjugate variable, measurement of which
means that the original state cannot be recovered. For example,
measurement of photon polarization at 45 degrees will quantum erase a
horizontal/vertical polarization state.

Going back to the original buckyball experiment, the balls are laser
heated and this puts them in an excited state. The state then decays
by IR emission. This happens

Re: Why are laws of physics stable?


On 12-07-2021 00:43, Bruce Kellett wrote:

On Mon, Jul 12, 2021 at 3:44 AM smitra  wrote:


On 11-07-2021 02:46, Bruce Kellett wrote:

On Sun, Jul 11, 2021 at 10:21 AM smitra  wrote:



This is only true in practice, not in principle because the

escaping


photons be captured and detected in principle.


Not true. You can;t ever catch up with the escaping photons to

capture them.



They can be reflected back using mirrors.


You can't decide to put mirrors in place after the photons have been
emitted.
Of course, you could always decide in advance to perform all
experiments in reflecting boxes, but that is not what is done in
practice. So the general rule is that escaping photons are not
recoverable.


If in one experiment with mirrors they are recoverable and therefore 
interference can be detected in principle, then that proves that the 
other sectors do objectively exist. Since the objective existence of 
these sectors does not depend on whether or not one uses mirrors in the 
experiment, one can conclude that they exist in general.






One can therefore at least in
principle do experiments that rule out collapse theories for large
systems that are large enough to include observers. And once
collapse
theories are ruled out in a few experiments it follows that in a
superposition all the sectors where observers find different results

objectively exist, regardless of whether or not in those particular
situations one could have done the same sorts of experiments that
ruled
out collapse theories.




It's also irrelevant,
because the photons that escape continue to exist, the

information

contained in  the photons continues to exist, even if it were

true that

they could not be recovered. Then if we were to conduct an
interference experiment with the balls then we wouldn't see an

interference

pattern. But if we write down where each balls lands on the

screen then this

information together with information that could be obtained by
performing certain measurements on the escaping photons emitted

by

each ball would still yield an interference pattern.


This is where you go seriously wrong. Simply recording where the
photons land does not quantum erase the information they carried.

Once

the photons carry off the which way information, the interference
pattern is restored only if the information carried by the photons

is

quantum erased. Simply running the photons into a screen (or the
wall), even if you record where they land, is not quantum erasure.
See, for example, the paper arXiv:1206.6578 on quantum erasure. In
this paper they say "the presence of path information anywhere in

the

universe is sufficient to prohibit any possibility of

interference. In

other words, the atoms' path states alone are not in a coherent
superposition due to the atom-photon entanglement." This transfers
directly to the buckyball experiment under discussion. Running the
photons into a screen, or the wall, does not destroy the

ball-photon

entanglement.



Recording where the photons land on the screen is enough, this is
very
easy to see. Let's consider an interference experiment where the
particle gets entangled with another particle that carries away the
which way information. If we work in the position representation
then we
have a wavefunction:

psi(x,y) = 1/sqrt(2) [psi_1(x,y) + psi_2(x,y)](1)

where x is the position of particle 1 just before it hits the screen
and
y the position of the other particle at that time, and psi_1(x,y) is
the
wavefunction when only slit 1 is open while psi_2(x,y) the
wavefunction
with only slit 2 open. We also assume that particle 2 carries the
which
way information, this means that psi_1(x,y) and psi_2(x,y) for x
kept
fixed and considered as a function of y are eigenfunctions with
different eigenvalues of an observable that corresponds to
extracting
the which way information from the second particle. This then
implies
that psi_1(x,y) and psi_2(x,y) are orthogonal for every x:

Integral psi_1(x,y)*psi_2(x,y)d^3y = 0 (2)


This is the same mistake that you have routinely made in all of these
discussions. You calculate for particles through each slit
independently.

No that's not what I did.

 But if that is the case, then no interference can ever

be seen. The possibility of interference relies on coherence between
the amplitudes at the two slits. In your calculations, you have never
allowed for this coherence -- the amplitudes at the two slits are not
independent.

No, I really think you need to refresh you knowledge of quantum 
mechanics.



Bruce


So, if we compute the probability of observing particle 1 at some
position x, while we don't measure the position of particle 2, we
find
by taking the modulus squared of (1) and integrating over y:

Integral |psi(x,y)|^2 d^3y = 1/2 Integral |psi_1(x,y)|^2 d^3y + 1/2
Integral |psi_2(x,y)|^2 d^3y

+ Re Integral psi_1(x,y)*psi_2(x,y) d^3y

And the last term vanishes due to (2), so we don't detect
interference.
But now suppose

Re: Why are laws of physics stable?

On Mon, Jul 12, 2021 at 8:43 AM Bruce Kellett  wrote:

> On Mon, Jul 12, 2021 at 3:44 AM smitra  wrote:
>
>> On 11-07-2021 02:46, Bruce Kellett wrote:
>>
> >
>> > This is where you go seriously wrong. Simply recording where the
>> > photons land does not quantum erase the information they carried. Once
>> > the photons carry off the which way information, the interference
>> > pattern is restored only if the information carried by the photons is
>> > quantum erased. Simply running the photons into a screen (or the
>> > wall), even if you record where they land, is not quantum erasure.
>> > See, for example, the paper arXiv:1206.6578 on quantum erasure. In
>> > this paper they say "the presence of path information anywhere in the
>> > universe is sufficient to prohibit any possibility of interference. In
>> > other words, the atoms' path states alone are not in a coherent
>> > superposition due to the atom-photon entanglement." This transfers
>> > directly to the buckyball experiment under discussion. Running the
>> > photons into a screen, or the wall, does not destroy the ball-photon
>> > entanglement.
>> >
>>
>> Recording where the photons land on the screen is enough, this is very
>> easy to see. Let's consider an interference experiment where the
>> particle gets entangled with another particle that carries away the
>> which way information. If we work in the position representation then we
>> have a wavefunction:
>>
>> psi(x,y) = 1/sqrt(2) [psi_1(x,y) + psi_2(x,y)](1)
>>
>> where x is the position of particle 1 just before it hits the screen and
>> y the position of the other particle at that time, and psi_1(x,y) is the
>> wavefunction when only slit 1 is open while psi_2(x,y) the wavefunction
>> with only slit 2 open. We also assume that particle 2 carries the which
>> way information, this means that psi_1(x,y) and psi_2(x,y) for x kept
>> fixed and considered as a function of y are eigenfunctions with
>> different eigenvalues of an observable that corresponds to extracting
>> the which way information from the second particle. This then implies
>> that psi_1(x,y) and psi_2(x,y) are orthogonal for every x:
>>
>> Integral psi_1(x,y)*psi_2(x,y)d^3y = 0 (2)
>>
>
>
> This is the same mistake that you have routinely made in all of these
> discussions. You calculate for particles through each slit independently.
> But if that is the case, then no interference can ever be seen. The
> possibility of interference relies on coherence between the amplitudes at
> the two slits. In your calculations, you have never allowed for this
> coherence -- the amplitudes at the two slits are not independent.
>

To see this in more detail, we can go through your original argument.
You write : denote the buckyball moving through the left slit by |L> and
the right slit by |R> and the photons by |PL> and |PR>. Now the states |R>
and |L> are coherent; thus correlated; and the overlap  does not
vanish. Writing

 |psi> = (|L>|PL>  +  |R>|PR>)

is the same as assuming that the photons are emitted from the slits if they
detect a ball passing through that slit. This is a direct "which-way"
measurement, and since the photon states are independent, they are, as you
say, orthogonal, and there can be no interference if such measurements are
made at the slits. This is the essential loss of coherence.

Your claim then is that if we observe the photons on another screen and
keep track of the ball positions on their screen for photons that arrive at
the same point on this other screen, then the interference pattern is
restored. This is wrong, because the photons emitted from the slits if and
when they detect a ball passing through the corresponding slit are still
independent. The fact that they might arrive at the same point on a further
screen is irrelevant. The photon states are still independent. In fact,
there can only ever be one photon for each buckyball, so there can be no
case in which two photons arrive at the photon screen at the same time. So
the photon states are necessarily always independent and orthogonal, and
your overlap function

   = 0,  always.

Consequently, your joint detection of photons does not restore the
interference. Decoherence, once present, destroys coherence. And this
coherence cannot be restored simply by observing the which-way photons. You
have to quantum erase the which-way information. And this is not easy for
this particular set-up. Quantum erasure essentially requires a measurement
of a conjugate variable, measurement of which means that the original state
cannot be recovered. For example, measurement of photon polarization at 45
degrees will quantum erase a horizontal/vertical polarization state.

Going back to the original buckyball experiment, the balls are laser heated
and this puts them in an excited state. The state then decays by IR
emission. This happens after the balls have passed the initial slits, so it
is not a direct which-way measurement. It is only as the temperature

Re: Why are laws of physics stable?

On Mon, Jul 12, 2021 at 3:44 AM smitra  wrote:

> On 11-07-2021 02:46, Bruce Kellett wrote:
> > On Sun, Jul 11, 2021 at 10:21 AM smitra  wrote:
> >>>
> >> This is only true in practice, not in principle because the escaping
> >>
> >> photons be captured and detected in principle.
> >
> > Not true. You can;t ever catch up with the escaping photons to
> capture them.
>


> They can be reflected back using mirrors.


You can't decide to put mirrors in place after the photons have been
emitted.
Of course, you could always decide in advance to perform all experiments in
reflecting boxes, but that is not what is done in practice. So the general
rule is that escaping photons are not recoverable.




> One can therefore at least in
> principle do experiments that rule out collapse theories for large
> systems that are large enough to include observers. And once collapse
> theories are ruled out in a few experiments it follows that in a
> superposition all the sectors where observers find different results
> objectively exist, regardless of whether or not in those particular
> situations one could have done the same sorts of experiments that ruled
> out collapse theories.
>
>
> >
> >> It's also irrelevant,
> >> because the photons that escape continue to exist, the information
> >> contained in  the photons continues to exist, even if it were true that
> >> they could not be recovered. Then if we were to conduct an
> >> interference experiment with the balls then we wouldn't see an
> interference
> >> pattern. But if we write down where each balls lands on the screen then
> this
> >> information together with information that could be obtained by
> >> performing certain measurements on the escaping photons emitted by
> >> each ball would still yield an interference pattern.
> >
> > This is where you go seriously wrong. Simply recording where the
> > photons land does not quantum erase the information they carried. Once
> > the photons carry off the which way information, the interference
> > pattern is restored only if the information carried by the photons is
> > quantum erased. Simply running the photons into a screen (or the
> > wall), even if you record where they land, is not quantum erasure.
> > See, for example, the paper arXiv:1206.6578 on quantum erasure. In
> > this paper they say "the presence of path information anywhere in the
> > universe is sufficient to prohibit any possibility of interference. In
> > other words, the atoms' path states alone are not in a coherent
> > superposition due to the atom-photon entanglement." This transfers
> > directly to the buckyball experiment under discussion. Running the
> > photons into a screen, or the wall, does not destroy the ball-photon
> > entanglement.
> >
>
> Recording where the photons land on the screen is enough, this is very
> easy to see. Let's consider an interference experiment where the
> particle gets entangled with another particle that carries away the
> which way information. If we work in the position representation then we
> have a wavefunction:
>
> psi(x,y) = 1/sqrt(2) [psi_1(x,y) + psi_2(x,y)](1)
>
> where x is the position of particle 1 just before it hits the screen and
> y the position of the other particle at that time, and psi_1(x,y) is the
> wavefunction when only slit 1 is open while psi_2(x,y) the wavefunction
> with only slit 2 open. We also assume that particle 2 carries the which
> way information, this means that psi_1(x,y) and psi_2(x,y) for x kept
> fixed and considered as a function of y are eigenfunctions with
> different eigenvalues of an observable that corresponds to extracting
> the which way information from the second particle. This then implies
> that psi_1(x,y) and psi_2(x,y) are orthogonal for every x:
>
> Integral psi_1(x,y)*psi_2(x,y)d^3y = 0 (2)
>


This is the same mistake that you have routinely made in all of these
discussions. You calculate for particles through each slit independently.
But if that is the case, then no interference can ever be seen. The
possibility of interference relies on coherence between the amplitudes at
the two slits. In your calculations, you have never allowed for this
coherence -- the amplitudes at the two slits are not independent.

Bruce


So, if we compute the probability of observing particle 1 at some
> position x, while we don't measure the position of particle 2, we find
> by taking the modulus squared of (1) and integrating over y:
>
> Integral |psi(x,y)|^2 d^3y = 1/2 Integral |psi_1(x,y)|^2 d^3y + 1/2
> Integral |psi_2(x,y)|^2 d^3y
>
>   + Re Integral psi_1(x,y)*psi_2(x,y) d^3y
>
> And the last term vanishes due to (2), so we don't detect interference.
> But now suppose that we perform a simultaneous measurement of the
> positions of both particles. The probability of detecting particle 1 at
> position x and particle 2 at position y is:
>
> |psi(x,y)|^2 = 1/2 |psi_1(x,y)|^2 y + 1/2 |psi_2(x,y)|^2  + Re
> psi_1(x,y)*psi_2(x,y)
>
> So, there now clearly an interference

Re: Why are laws of physics stable?

On Mon, Jul 12, 2021 at 7:13 AM 'Brent Meeker' via Everything List <
everything-list@googlegroups.com> wrote:

> On 7/11/2021 9:53 AM, smitra wrote:
> > On 11-07-2021 02:41, 'Brent Meeker' via Everything List wrote:
> >> On 7/10/2021 5:35 PM, smitra wrote:
> >>> On 11-07-2021 01:05, Bruce Kellett wrote:
>  On Sun, Jul 11, 2021 at 1:18 AM smitra  wrote:
> 
> > On 10-07-2021 07:58, Bruce Kellett wrote:
> >>
> >> Neither of your analyses actually explain the observed behaviour.
> >
> > It does so, otherwise the experiment would have proven that QM is
> > invalid.
> 
>  So, in your opinion, the quantum mechanical analysis given in the
>  paper is wrong.
> >>>
> >>> It is correct, it's just that your conclusion about the MWI based on
> >>> that paper is incorrect. Escaping infrared photons and the
> >>> decoherence this causes are totally irrelevant. Whether or not an
> >>> interference pattern can be detected is not relevant to the question
> >>> of whether or not a superposition exists when we know that it exist
> >>> and it has no decohered. It's just that you then can't reproduce one
> >>> particular line of evidence for the validity of quantum mechanics in
> >>> that particular experiment.
> >>>
> >>> If I prepare the state of a particle in a  superposition and let
> >>> this interact in a certain way, then we know how this state will
> >>> evolve. If this evolution involves interactions with many particles,
> >>> then the system will decohere. Then it may be true that we cannot
> >>> distinguish the state of the system from being in a pure or mixed
> >>> state in practice due to not being able to conduct an interference
> >>> experiment involving a very large number of particles, but quantum
> >>> mechanics still tells us that  the superposition exists and that if
> >>> we were to conduct the right sort of interference experiment, we
> >>> would see an interference that would prove that the state is not a
> >>> mixed state.
> >>
> >> But does it exist if part of the information has crossed the Hubble
> >> boundary?  Then there is no experiment, even in principle, that would
> >> prove the state is not a mixed one.
> >>
> >
> > The theories that tell you that there is such a thing like a Hubble
> > boundary, also tell you that information does not vanish. We can now
> > detect photons from galaxies that have always receded from us faster
> > than light.
>
> ?? I think that's a contradiction.
>

It's worse than that -- it's gibberish.

> > So, that's information we're that has leaked away from those galaxies
> > that does objectively exists. To prove that a state is a pure state
> > one can just point to the way the state was prepared and invoke
> > quantum mechanics. It's not necessary to demonstrate interference to
> > prove that quantum mechanics is still valid.
>
> So, then you agree with Bruce that so long as the which-way information
> exists in those outgoing photons, the interference pattern will show up?
>

I think you might have meant that as long as the which-way information
exists (has not been quantum erased), then no interference is visible.

Bruce

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Re: Why are laws of physics stable?

2021-07-11 Thread 'Brent Meeker' via Everything List

On 7/11/2021 9:53 AM, smitra wrote:

On 11-07-2021 02:41, 'Brent Meeker' via Everything List wrote:

On 7/10/2021 5:35 PM, smitra wrote:

On 11-07-2021 01:05, Bruce Kellett wrote:

On Sun, Jul 11, 2021 at 1:18 AM smitra wrote:

On 10-07-2021 07:58, Bruce Kellett wrote:

Neither of your analyses actually explain the observed behaviour.

It does so, otherwise the experiment would have proven that QM is
invalid.

So, in your opinion, the quantum mechanical analysis given in the
paper is wrong.

It is correct, it's just that your conclusion about the MWI based on
that paper is incorrect. Escaping infrared photons and the
decoherence this causes are totally irrelevant. Whether or not an
interference pattern can be detected is not relevant to the question
of whether or not a superposition exists when we know that it exist
and it has no decohered. It's just that you then can't reproduce one
particular line of evidence for the validity of quantum mechanics in
that particular experiment.

If I prepare the state of a particle in a superposition and let
this interact in a certain way, then we know how this state will
evolve. If this evolution involves interactions with many particles,
then the system will decohere. Then it may be true that we cannot
distinguish the state of the system from being in a pure or mixed
state in practice due to not being able to conduct an interference
experiment involving a very large number of particles, but quantum
mechanics still tells us that the superposition exists and that if
we were to conduct the right sort of interference experiment, we
would see an interference that would prove that the state is not a
mixed state.

But does it exist if part of the information has crossed the Hubble
boundary? Then there is no experiment, even in principle, that would
prove the state is not a mixed one.

?? I think that's a contradiction.

So, that's information we're that has leaked away from those galaxies
that does objectively exists. To prove that a state is a pure state
one can just point to the way the state was prepared and invoke
quantum mechanics. It's not necessary to demonstrate interference to
prove that quantum mechanics is still valid.

So, then you agree with Bruce that so long as the which-way information
exists in those outgoing photons, the interference pattern will show up?

Brent

Saibal

Brent

Re: Why are laws of physics stable?

2021-07-11 Thread 'Brent Meeker' via Everything List

On 7/11/2021 5:08 AM, John Clark wrote:
Brent Meeker' via Everything List > wrote:

> /As I recall even Shor's algorithm has a probabilistic step so
that the answer is only correct with high probability, not
certainty. /

So your argument is that a Quantum Computer can't be intelligent
because it is capable of committing error.

That's not only not my argument it's not my opinion either. From where
to you pull these assumptions?

Well,.. a human brain is most certainly capable of committing error,
in fact it's very very good at doing exactly that, and yet it seems to
be capable of intelligent behavior, at least some of the time.

/> I don't think a "being" can have knowledge that can be quantum
erased. Such a "being" would have to be isolated from the
environment /

A quantum computer, or a computer of any sort for that matter, that is
totally isolated from the environment would be absolutely useless.
Both a quantum computer and a human brain must be isolated to some
degree, that's why we have a bone skull; admittedly for a Quantum
Computer the isolation must be more sophisticated and extensive so
that the only changes made from the outside come from deliberate
changes carefully made by keyboards and other precision input devices.

/> and have a relatively small number of degrees of freedom. /

That is nonsense, the entire advantage of Quantum Computers is that
they have vastly more degrees of freedom than a conventional computer.
To describe the state of a n bit conventional processor you'd need n
real numbers; but to describe the state of a n qubit Quantum Computer
you'd need 2^n complex numbers, and thanks to the Born rule you need 2
complex numbers to define a unique real number so that means a n qubit
Quantum Computer has 2*(2^n) -2 real degrees of freedom. The -2 is
there because you have to remove 2 to normalized phase and amplitude.

You seemed to have missed the conditional "...that can be quantum erased".

Brent

/> you're never going to find a being that behaves intelligently
based on information that can be quantum erased./

If that is true it would mean there's something in the fundamental
laws of physics that would preventthe construction of large scale
Quantum Computers, as we already know it's possible to make small
Quantum Computers because we've already done it. Quantum Computer
expert Scott Aaronsonhas said that if somebody could demonstrate that
that it's true they can't be made then for him personally it would be
even more exciting then if somebody actually made a large scale
Quantum Computer because he is a theoretician, and as the fundamental
laws of physics are currently understood a large scale quantum
computer is possible, so if it's proved it's not possible after all
then that could only mean new laws of fundamental physics have been
discovered. To insist they can't be made you must invoke new
hypothetical physics.

John K Clark See what's on my new list at Extropolis

qqnn

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Re: Why are laws of physics stable?

2021-07-11 Thread 'Brent Meeker' via Everything List

On 7/11/2021 4:03 AM, Tomas Pales wrote:

On Saturday, July 10, 2021 at 11:34:28 PM UTC+2 Brent wrote:

I think this kind of talk puts far too much on consciousness.
Conscious thoughts seem to pop into my head with no antecedents,
yet they relate to past and distant things in my experience. The
Poincare' effect shows that even the most abstract thought is
largely unconscious.

What is Poincare effect?

https://khuntersscience.blogspot.com/2012/09/an-essay-on-mathematical-creation-by.html

Consciousness seems to be the necessary basis of personal identity: "I
am conscious therefore I am". Of course it depends on unconscious
parts of the brain, the rest of the body and the environment; where to
draw the boundary of personal identity is somewhat blurry.

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Re: Why are laws of physics stable?

2021-07-11 Thread spudboy100 via Everything List

I am guessing that the reason physics is stable or the mechanism for this is 
the quantum weirdness observations of old as taught by Schrodinger, Heisenberg, 
Wigner, Wheeler, and Bell, all typified by repeated Stern - Gerlach 
experiments. 
https://www.forbes.com/sites/startswithabang/2021/07/09/ask-ethan-is-there-a-hidden-quantum-reality-underlying-what-we-observe/?sh=43131f4fef72Somehow
 reality is crystalized by some "observer" and things may be programmed to 
generate a result. This doesn't answer the next question but it is a summary 
how the cosmos behaves. Why it is programmed in such a manner or how it is open 
to conjecture. This question needs a big project budget.

-Original Message-
From: Jason Resch 
To: Everything List 
Sent: Sat, Jul 10, 2021 7:01 am
Subject: Re: Why are laws of physics stable?

On Sat, Jul 10, 2021, 1:58 AM Bruce Kellett  wrote:

On Sat, Jul 10, 2021 at 10:21 AM smitra  wrote:

On 08-07-2021 01:51, Bruce Kellett wrote:
> 
> Do you dispute that that is what the paper by Hornberger et al. says?
> 
> Bruce

I don't dispute these results. The buckyballs are coming from a thermal 
reservoir at some finite temperature. We can avoid working with mixed 
states by simply considering the interference pattern for each pure 
state separately and then summing over the probability distribution over 
the pure states. But for this discussion we want to focus on what the 
interference pattern will be if all the buckyballs are in the same 
exited state. If we put the entire system ina finite volume then we have 
a countable set of allowed k-values for the photon momenta. We then have 
a set of allowed states for the photons defined by the allowed momenta 
and a polarization. We can then label these photon states using a number 
and then specify an arbitrary state for the photons by specifying how 
many photons we have in each state, and these numbers can then be equal 
to zero. If they are all zero then no photons are present.

If only the right slit is open, the state of the buckyball and the 
photons just before the screen is hit can be denoted as:

|Right> = sum over n1, n2,n3,...|R(n1,n2,n3,...)>|n1,n2,n3,..>

where |R(n1,n2,...)> denotes the quantum state of the buckyball if it 
emits n1 photons in state 1, n2 photons in state 2 etc. The state of the 
photons is then denoted as |n1,n2,n3,..>

If only the left slit is open, the state of the buckyball and the 
photons just before the screen is hit can be denoted as:

|Left> = sum over n1, n2,n3,...|L(n1,n2,n3,...)>|n1,n2,n3,..>

where |R(n1,n2,...)> denotes the quantum state of the buckyball. Here we 
note that the state of the photons will pick up a phase factor relative 
to the case of only the right slit being open, but we can then absorb 
this phase factor in |L(n1,n2,n3,...)>.

With both slits open, we'll then have a state of the form:

|psi> = 1/sqrt(2) [|Right> + |Left>]

The inner product of |psi> with some position eigenstate |x>,  is 
then a state vector for the photon states, the squared norm of that 
state vector is the probability of finding the buckyball at position x, 
because this is the sum over the probabilities for photons over all 
possible photon states. So, the probability is then:

P(x) = ^2 = 1/2 [^2 + ^2] + 
Re[]

We can then evaluate the interference term as follows:

Re[] = Re sum over n1, n2,n3,...m1,m2,m3 
*

Using that  = 0 unless m_j = n_j for all 
j, in which case this inner product equals 1, we then have:

Re[] = Re sum over n1, n2,n3 
* =

Re[*] + 
Re[*] + .  (1)

As explained above when there are photons present then we've absorbed 
the phase factor due to translation of the photon states in the states 
|L(n1,n2,n3...)>. For each wave vector k there is a factor for each 
photon with that wavevector of exp(i k dot r) where r is the position of 
the left slit w.r.t. the right slit. So, the total phase factor will be:

Product over j of exp(i kj nj dot r)

In the experiment there is then an additional summation over the pure 
states of the buckyballs. If the temperature is low then the summation 
will consist of states for which |R(0,0,0,..> and |L(0,0,0,..> are the 
dominant terms, as most of the time no photons will be emitted. At 
higher temperatures the typical states there will be contributions from 
different numbers pf photons, so the interference pattern will be a sum 
of many different terms in (1) with comparable norms, they come with 
different phase factors due to the different numbers of photons with 
different momenta. So, the interference pattern will be washed out.

This analysis contradicts what you said in your first analysis. In the first 
analysis, you claimed that no interference would be seen if the IR photons were 
not detected. You seem to have dropped this notion in the above. You now say 
that there is interference when no photons are present, but this is was

Re: Why are laws of physics stable?


On 11-07-2021 02:46, Bruce Kellett wrote:

On Sun, Jul 11, 2021 at 10:21 AM smitra  wrote:


On 11-07-2021 00:59, Bruce Kellett wrote:

On Sun, Jul 11, 2021 at 8:27 AM smitra  wrote:


On 10-07-2021 21:41, 'Brent Meeker' via Everything List wrote:

On 7/10/2021 1:19 AM, Bruno Marchal wrote:

So, in general we can conclude by generalizing this to any

large

number of particles that even with what we consider to be

permanent

records, you don't get rid of the theoretical possibility of
interference between the sectors where those records are

different.


We can if the universe is expanding faster than light beyond the

Hubble

radius.



The expansion of the universe is irrelevant. The information

needed

to see the interference pattern continues to exist outside the

horizon

when it isn't accessible to us anymore. And this is irrelevant

for the

discussions about observations in quantum mechanics. If an

observer

performs a measurement and the claim is that this is a unitary
process with the observer evolving into a superposition, while

the objection


against this claim is that infrared photons are escaping and will
eventually move beyond the Hubble volume, then these photons will
still not have escaped beyond the Hubble horizon by the time the

observer

is aware of the results of the experiment. So, whether or not the
photons will eventually no longer be accessible, cannot be

relevant.


Once the photons escape from the immediate environs of the

experiment,

they are not recoverable. Try shining a torch at night to

illuminate a

tree. Now try to stop the illumination already present. You can

stop

future illumination by covering the torch, or switching it off.

But

once the tree is illuminated it is not reversible. The expansion

of

the universe, and the existence of the Hubble horizon, just makes

the

irreversibility more obvious.


This is only true in practice, not in principle because the escaping

photons be captured and detected in principle.


Not true. You can;t ever catch up with the escaping photons to capture
them.
They can be reflected back using mirrors. One can therefore at least in 
principle do experiments that rule out collapse theories for large 
systems that are large enough to include observers. And once collapse 
theories are ruled out in a few experiments it follows that in a 
superposition all the sectors where observers find different results 
objectively exist, regardless of whether or not in those particular 
situations one could have done the same sorts of experiments that ruled 
out collapse theories.






It's also irrelevant,
because the photons that escape continue to exist, the information
contained in  the photons continues to exist, even if it were true
that
they could not be recovered. Then if we were to conduct an
interference
experiment with the balls then we wouldn't see an interference
pattern.
But if we write down where each balls lands on the screen then this
information together with information that could be obtained by
performing certain measurements on the escaping photons emitted by
each
ball would still yield an interference pattern.


This is where you go seriously wrong. Simply recording where the
photons land does not quantum erase the information they carried. Once
the photons carry off the which way information, the interference
pattern is restored only if the information carried by the photons is
quantum erased. Simply running the photons into a screen (or the
wall), even if you record where they land, is not quantum erasure.
See, for example, the paper arXiv:1206.6578 on quantum erasure. In
this paper they say "the presence of path information anywhere in the
universe is sufficient to prohibit any possibility of interference. In
other words, the atoms' path states alone are not in a coherent
superposition due to the atom-photon entanglement." This transfers
directly to the buckyball experiment under discussion. Running the
photons into a screen, or the wall, does not destroy the ball-photon
entanglement.



Recording where the photons land on the screen is enough, this is very 
easy to see. Let's consider an interference experiment where the 
particle gets entangled with another particle that carries away the 
which way information. If we work in the position representation then we 
have a wavefunction:


psi(x,y) = 1/sqrt(2) [psi_1(x,y) + psi_2(x,y)](1)

where x is the position of particle 1 just before it hits the screen and 
y the position of the other particle at that time, and psi_1(x,y) is the 
wavefunction when only slit 1 is open while psi_2(x,y) the wavefunction 
with only slit 2 open. We also assume that particle 2 carries the which 
way information, this means that psi_1(x,y) and psi_2(x,y) for x kept 
fixed and considered as a function of y are eigenfunctions with 
different eigenvalues of an observable that corresponds to extracting 
the which way information from the second particle. This then implies 
that

RE: Why are laws of physics stable?

2021-07-11 Thread Philip Benjamin

[Philip Benjamin]
 Confusion worse confounded?  Permanent records? Theoretical possibility of 
different sectors interfering? Quarks and electrons are different sectors with 
"permanent records". Will they not just "whizz around", at least say at some 
"beginnings" (the imaginary Big Bang). How do they eventually form "protons" 
and "neutrons" and remain separate from "electrons"? How do these "protons" 
(and neutrons) remain "bound" together to form a nucleus of an atom, with 
electrons in "permanent records of energy levels" orbiting around but not 
falling into the nucleus? From whence comes the "permanent record" that an atom 
may combine with another atom according to "permant records of the Laws of 
Chemistry? From whence comes the "permanent record" that there will be atoms of 
inert gases with a "stable" duet or octet of "electrons"? Will there be any 
universe at all without these "permanent records", CopenPagan Interpretation 
notwithstanding?
   Philip Benjamin  

From: 'Brent Meeker' via Everything List  
Sent: Saturday, July 10, 2021 2:41 PM  Subject: Re: Why are laws of physics 
stable?

On 7/10/2021 1:19 AM, Bruno Marchal wrote:
> So, in general we can conclude by generalizing this to any large number of 
> particles that even with what we consider to be permanent records, you don't 
> get rid of the theoretical possibility of interference between the sectors 
> where those records are different. We can if the universe is expanding faster 
> than light beyond the Hubble radius.

Brent
.

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Re: Why are laws of physics stable?

On 11-07-2021 02:57, Bruce Kellett wrote:

On Sun, Jul 11, 2021 at 10:36 AM smitra wrote:

On 11-07-2021 01:05, Bruce Kellett wrote:

On Sun, Jul 11, 2021 at 1:18 AM smitra wrote:

On 10-07-2021 07:58, Bruce Kellett wrote:

Neither of your analyses actually explain the observed

behaviour.

It does so, otherwise the experiment would have proven that QM is
invalid.

So, in your opinion, the quantum mechanical analysis given in the
paper is wrong.

It is correct, it's just that your conclusion about the MWI based on

that paper is incorrect. Escaping infrared photons and the
decoherence
this causes are totally irrelevant. Whether or not an interference
pattern can be detected is not relevant to the question of whether
or
not a superposition exists when we know that it exist and it has no
decohered.

The question is not whether or not there is a superposition. The
question is whether or not the superposition is coherent. If the balls
are entangled with photons that have escaped, the superposition has
decohered and is no longer coherent. Loss of coherence means that
there can be no interference pattern. It is not just that it is not
visible -- it simply no longer exists. Coherence can only be restored
if the escaping information is quantum erased. The escaping photons
are part of the total system. If you ignore this entanglement, the
remaining system is a mixed state.

Yes, so you need a combined measured on the balls and the photons to
demonstrate interference. But whether or not we actually do this is then
irrelevant to the question of whether or not in a superposition both
sectors of that superposition actually exist, unless one would doubt the
validity of quantum mechanics and would want to test the quantum
mechanical prediction. If we assume that quantum mechanics is valid then
we assume that both sectors objectively exist, regardless of whether or
not you could actually perform the required interference experiment.

Saibal

Bruce

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Re: Why are laws of physics stable?

On 11-07-2021 00:59, Bruce Kellett wrote:

On Sun, Jul 11, 2021 at 8:27 AM smitra wrote:

On 10-07-2021 21:41, 'Brent Meeker' via Everything List wrote:

On 7/10/2021 1:19 AM, Bruno Marchal wrote:

So, in general we can conclude by generalizing this to any large
number of particles that even with what we consider to be

permanent

records, you don't get rid of the theoretical possibility of
interference between the sectors where those records are

different.

We can if the universe is expanding faster than light beyond the

Hubble

radius.

against this claim is that infrared photons are escaping and will
eventually move beyond the Hubble volume, then these photons will
still
not have escaped beyond the Hubble horizon by the time the observer
is
aware of the results of the experiment. So, whether or not the
photons
will eventually no longer be accessible, cannot be relevant.

Once the photons escape from the immediate environs of the experiment,
they are not recoverable. Try shining a torch at night to illuminate a
tree. Now try to stop the illumination already present. You can stop
future illumination by covering the torch, or switching it off. But
once the tree is illuminated it is not reversible. The expansion of
the universe, and the existence of the Hubble horizon, just makes the
irreversibility more obvious.

Nothing stops you from doing the experiment in a closed volume with
mirrors at the boundary. Since what we measure in a experiments is
determined by the laws of physics which are local in nature, the sort of
we impose boundary conditions on photons far away is irrelevant. Whether
or not an observer measuring the z-component of a spin that was
polarized in the positive x-direction ends up in a superposition where
there are objectively both outcomes being realized or only one outcome
does not depend on whether or not all the escaping infrared photons can
actually be captured and be made part of a giant interference experiment
to demonstrate an interference to prove that the superposition indeed
exists.

The existence of the superposition depends on the validity of quantum
mechanics for large systems. One can prove or disprove that in suitable
experiments. Once that's done and, say, collapse theories are ruled out
then it's not necessary to prove this over and over again for each and
every case. As things stand now, we can't rule out that an objective
collapse can happen, but this would require a violation of quantum
mechanics.

Bruce

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Re: Why are laws of physics stable?

On 11-07-2021 02:41, 'Brent Meeker' via Everything List wrote:

On 7/10/2021 5:35 PM, smitra wrote:

On 11-07-2021 01:05, Bruce Kellett wrote:

On Sun, Jul 11, 2021 at 1:18 AM smitra wrote:

On 10-07-2021 07:58, Bruce Kellett wrote:

Neither of your analyses actually explain the observed behaviour.

It does so, otherwise the experiment would have proven that QM is
invalid.

So, in your opinion, the quantum mechanical analysis given in the
paper is wrong.

It is correct, it's just that your conclusion about the MWI based on
that paper is incorrect. Escaping infrared photons and the decoherence
this causes are totally irrelevant. Whether or not an interference
pattern can be detected is not relevant to the question of whether or
not a superposition exists when we know that it exist and it has no
decohered. It's just that you then can't reproduce one particular line
of evidence for the validity of quantum mechanics in that particular
experiment.

If I prepare the state of a particle in a superposition and let this
interact in a certain way, then we know how this state will evolve. If
this evolution involves interactions with many particles, then the
system will decohere. Then it may be true that we cannot distinguish
the state of the system from being in a pure or mixed state in
practice due to not being able to conduct an interference experiment
involving a very large number of particles, but quantum mechanics
still tells us that the superposition exists and that if we were to
conduct the right sort of interference experiment, we would see an
interference that would prove that the state is not a mixed state.

But does it exist if part of the information has crossed the Hubble
boundary? Then there is no experiment, even in principle, that would
prove the state is not a mixed one.

The theories that tell you that there is such a thing like a Hubble
boundary, also tell you that information does not vanish. We can now
detect photons from galaxies that have always receded from us faster
than light. So, that's information we're that has leaked away from those
galaxies that does objectively exists. To prove that a state is a pure
state one can just point to the way the state was prepared and invoke
quantum mechanics. It's not necessary to demonstrate interference to
prove that quantum mechanics is still valid.

Saibal

Brent

Re: Why are laws of physics stable?

2021-07-11 Thread John Clark

Brent Meeker' via Everything List  wrote:

 > *As I recall even Shor's algorithm has a probabilistic step so that the
> answer is only correct with high probability, not certainty. *
>

So your argument is that a Quantum Computer can't be intelligent because it
is capable of committing error. Well,... a human brain is most certainly
capable of committing error, in fact it's very very good at doing exactly
that, and yet it seems to be capable of intelligent behavior, at least some
of the time.

> *>  I don't think a "being" can have knowledge that can be quantum
> erased.  Such a "being" would have to be isolated from the environment *

A quantum computer, or a computer of any sort for that matter, that is
totally isolated from the environment would be absolutely useless. Both a
quantum computer and a human brain must be isolated to some degree, that's
why we have a bone skull; admittedly for a Quantum Computer the isolation
must be more sophisticated and extensive so that the only changes made from
the outside come from deliberate changes carefully made by keyboards and
other precision input devices.

*> and have a relatively small number of degrees of freedom. *

That is nonsense, the entire advantage of Quantum Computers is that they
have vastly more degrees of freedom than a conventional computer. To
describe the state of a n bit conventional processor you'd need n real
numbers; but to describe the state of a n qubit Quantum Computer you'd need
2^n complex numbers, and thanks to the Born rule you need 2 complex numbers
to define a unique real number so that means a n qubit Quantum Computer has
2*(2^n) -2 real degrees of freedom. The -2 is there because you have to
remove 2 to normalized phase and amplitude.

* > you're never going to find a being that behaves intelligently based on
> information that can be quantum erased.*

If that is true it would mean there's something in the fundamental laws of
physics that would prevent the construction of large scale Quantum
Computers, as we already know it's possible to make small Quantum Computers
because we've already done it. Quantum Computer expert Scott Aaronson has
said that if somebody could demonstrate that that it's true they can't be
made then for him personally it would be even more exciting then if
somebody actually made a large scale Quantum Computer because he is a
theoretician, and as the fundamental laws of physics are currently
understood a large scale quantum computer is possible, so if it's proved
it's not possible after all then that could only mean new laws of
fundamental physics have been discovered. To insist they can't be made you
must invoke new hypothetical physics.

John K ClarkSee what's on my new list at  Extropolis

qqnn

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Re: Why are laws of physics stable?

2021-07-11 Thread Tomas Pales

On Saturday, July 10, 2021 at 11:34:28 PM UTC+2 Brent wrote:

>
> I think this kind of talk puts far too much on consciousness.  Conscious 
> thoughts seem to pop into my head with no antecedents, yet they relate to 
> past and distant things in my experience.  The Poincare' effect shows that 
> even the most abstract thought is largely unconscious.
>

What is Poincare effect?

Consciousness seems to be the necessary basis of personal identity: "I am 
conscious therefore I am". Of course it depends on unconscious parts of the 
brain, the rest of the body and the environment; where to draw the boundary 
of personal identity is somewhat blurry.

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Re: Why are laws of physics stable?

On Sun, Jul 11, 2021 at 10:36 AM smitra  wrote:

> On 11-07-2021 01:05, Bruce Kellett wrote:
> > On Sun, Jul 11, 2021 at 1:18 AM smitra  wrote:
> >
> >> On 10-07-2021 07:58, Bruce Kellett wrote:
> >>>
> >>> Neither of your analyses actually explain the observed behaviour.
> >>
> >> It does so, otherwise the experiment would have proven that QM is
> >> invalid.
> >
> > So, in your opinion, the quantum mechanical analysis given in the
> > paper is wrong.
>
> It is correct, it's just that your conclusion about the MWI based on
> that paper is incorrect. Escaping infrared photons and the decoherence
> this causes are totally irrelevant. Whether or not an interference
> pattern can be detected is not relevant to the question of whether or
> not a superposition exists when we know that it exist and it has no
> decohered.

The question is not whether or not there is a superposition. The question
is whether or not the superposition is coherent. If the balls are entangled
with photons that have escaped, the superposition has decohered and is no
longer coherent. Loss of coherence means that there can be no interference
pattern. It is not just that it is not visible -- it simply no longer
exists. Coherence can only be restored if the escaping information is
quantum erased. The escaping photons are part of the total system. If you
ignore this entanglement, the remaining system is a mixed state.

Bruce

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Re: Why are laws of physics stable?

On Sun, Jul 11, 2021 at 10:21 AM smitra  wrote:

> On 11-07-2021 00:59, Bruce Kellett wrote:
> > On Sun, Jul 11, 2021 at 8:27 AM smitra  wrote:
> >
> >> On 10-07-2021 21:41, 'Brent Meeker' via Everything List wrote:
> >>> On 7/10/2021 1:19 AM, Bruno Marchal wrote:
>  So, in general we can conclude by generalizing this to any large
>  number of particles that even with what we consider to be
> >> permanent
>  records, you don't get rid of the theoretical possibility of
>  interference between the sectors where those records are
> >> different.
> >>>
> >>> We can if the universe is expanding faster than light beyond the Hubble
> >>> radius.
> >>>
> >>
> >> The expansion of the universe is irrelevant. The information needed
> >> to see the interference pattern continues to exist outside the horizon
> >> when it isn't accessible to us anymore. And this is irrelevant for the
> >> discussions about observations in quantum mechanics. If an observer
> >> performs a measurement and the claim is that this is a unitary
> >> process with the observer evolving into a superposition, while the
> objection
> >>
> >> against this claim is that infrared photons are escaping and will
> >> eventually move beyond the Hubble volume, then these photons will
> >> still not have escaped beyond the Hubble horizon by the time the
> observer
> >> is aware of the results of the experiment. So, whether or not the
> >> photons will eventually no longer be accessible, cannot be relevant.
> >
> > Once the photons escape from the immediate environs of the experiment,
> > they are not recoverable. Try shining a torch at night to illuminate a
> > tree. Now try to stop the illumination already present. You can stop
> > future illumination by covering the torch, or switching it off. But
> > once the tree is illuminated it is not reversible. The expansion of
> > the universe, and the existence of the Hubble horizon, just makes the
> > irreversibility more obvious.
> >
> This is only true in practice, not in principle because the escaping
> photons be captured and detected in principle.


Not true. You can;t ever catch up with the escaping photons to capture them.

It's also irrelevant,
> because the photons that escape continue to exist, the information
> contained in  the photons continues to exist, even if it were true that
> they could not be recovered. Then if we were to conduct an interference
> experiment with the balls then we wouldn't see an interference pattern.
> But if we write down where each balls lands on the screen then this
> information together with information that could be obtained by
> performing certain measurements on the escaping photons emitted by each
> ball would still yield an interference pattern.
>


This is where you go seriously wrong. Simply recording where the photons
land does not quantum erase the information they carried. Once the photons
carry off the which way information, the interference pattern is restored
only if the information carried by the photons is quantum erased. Simply
running the photons into a screen (or the wall), even if you record where
they land, is not quantum erasure. See, for example, the paper
arXiv:1206.6578 on quantum erasure. In this paper they say "the presence of
path information anywhere in the universe is sufficient to prohibit any
possibility of interference. In other words, the atoms' path states alone
are not in a coherent superposition due to the atom-photon entanglement."
This transfers directly to the buckyball experiment under discussion.
Running the photons into a screen, or the wall, does not destroy the
ball-photon entanglement.

Bruce

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Re: Why are laws of physics stable?

2021-07-10 Thread 'Brent Meeker' via Everything List

On 7/10/2021 5:35 PM, smitra wrote:

On 11-07-2021 01:05, Bruce Kellett wrote:

On Sun, Jul 11, 2021 at 1:18 AM smitra wrote:

On 10-07-2021 07:58, Bruce Kellett wrote:

Neither of your analyses actually explain the observed behaviour.

It does so, otherwise the experiment would have proven that QM is
invalid.

So, in your opinion, the quantum mechanical analysis given in the
paper is wrong.

It is correct, it's just that your conclusion about the MWI based on
that paper is incorrect. Escaping infrared photons and the decoherence
this causes are totally irrelevant. Whether or not an interference
pattern can be detected is not relevant to the question of whether or
not a superposition exists when we know that it exist and it has no
decohered. It's just that you then can't reproduce one particular line
of evidence for the validity of quantum mechanics in that particular
experiment.

If I prepare the state of a particle in a superposition and let this
interact in a certain way, then we know how this state will evolve. If
this evolution involves interactions with many particles, then the
system will decohere. Then it may be true that we cannot distinguish
the state of the system from being in a pure or mixed state in
practice due to not being able to conduct an interference experiment
involving a very large number of particles, but quantum mechanics
still tells us that the superposition exists and that if we were to
conduct the right sort of interference experiment, we would see an
interference that would prove that the state is not a mixed state.

But does it exist if part of the information has crossed the Hubble
boundary? Then there is no experiment, even in principle, that would
prove the state is not a mixed one.

Brent

Then given that we don't have any experimental evidence to doubt the
validity of quantum mechanics, saying that the superposition does not
really exists in such a a case, is just silly.

Saibal

Bruce

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Re: Why are laws of physics stable?

On 11-07-2021 01:05, Bruce Kellett wrote:

On Sun, Jul 11, 2021 at 1:18 AM smitra wrote:

On 10-07-2021 07:58, Bruce Kellett wrote:

Neither of your analyses actually explain the observed behaviour.

It does so, otherwise the experiment would have proven that QM is
invalid.

So, in your opinion, the quantum mechanical analysis given in the
paper is wrong.

It is correct, it's just that your conclusion about the MWI based on
that paper is incorrect. Escaping infrared photons and the decoherence
this causes are totally irrelevant. Whether or not an interference
pattern can be detected is not relevant to the question of whether or
not a superposition exists when we know that it exist and it has no
decohered. It's just that you then can't reproduce one particular line
of evidence for the validity of quantum mechanics in that particular
experiment.

If I prepare the state of a particle in a superposition and let this
interact in a certain way, then we know how this state will evolve. If
this evolution involves interactions with many particles, then the
system will decohere. Then it may be true that we cannot distinguish the
state of the system from being in a pure or mixed state in practice due
to not being able to conduct an interference experiment involving a very
large number of particles, but quantum mechanics still tells us that
the superposition exists and that if we were to conduct the right sort
of interference experiment, we would see an interference that would
prove that the state is not a mixed state.

Then given that we don't have any experimental evidence to doubt the
validity of quantum mechanics, saying that the superposition does not
really exists in such a a case, is just silly.

Saibal

Bruce

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Re: Why are laws of physics stable?

On 11-07-2021 00:59, Bruce Kellett wrote:

On Sun, Jul 11, 2021 at 8:27 AM smitra wrote:

On 10-07-2021 21:41, 'Brent Meeker' via Everything List wrote:

On 7/10/2021 1:19 AM, Bruno Marchal wrote:

So, in general we can conclude by generalizing this to any large
number of particles that even with what we consider to be

permanent

records, you don't get rid of the theoretical possibility of
interference between the sectors where those records are

different.

We can if the universe is expanding faster than light beyond the

Hubble

radius.

This is only true in practice, not in principle because the escaping
photons be captured and detected in principle. It's also irrelevant,
because the photons that escape continue to exist, the information
contained in the photons continues to exist, even if it were true that
they could not be recovered. Then if we were to conduct an interference
experiment with the balls then we wouldn't see an interference pattern.
But if we write down where each balls lands on the screen then this
information together with information that could be obtained by
performing certain measurements on the escaping photons emitted by each
ball would still yield an interference pattern.

Bruce

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Re: Why are laws of physics stable?

On Sun, Jul 11, 2021 at 1:18 AM smitra  wrote:

> On 10-07-2021 07:58, Bruce Kellett wrote:
> >
> > Neither of your analyses actually explain the observed behaviour.
>
> It does so, otherwise the experiment would have proven that QM is
> invalid.
>

So, in your opinion, the quantum mechanical analysis given in the paper is
wrong.

Bruce

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Re: Why are laws of physics stable?

On Sun, Jul 11, 2021 at 8:27 AM smitra  wrote:

> On 10-07-2021 21:41, 'Brent Meeker' via Everything List wrote:
> > On 7/10/2021 1:19 AM, Bruno Marchal wrote:
> >> So, in general we can conclude by generalizing this to any large
> >> number of particles that even with what we consider to be permanent
> >> records, you don't get rid of the theoretical possibility of
> >> interference between the sectors where those records are different.
> >
> > We can if the universe is expanding faster than light beyond the Hubble
> > radius.
> >
>
> The expansion of the universe is irrelevant. The information needed to
> see the interference pattern continues to exist outside the horizon when
> it isn't accessible to us anymore. And this is irrelevant for the
> discussions about observations in quantum mechanics. If an observer
> performs a measurement and the claim is that this is a unitary process
> with the observer evolving into a superposition, while the objection
> against this claim is that infrared photons are escaping and will
> eventually move beyond the Hubble volume, then these photons will still
> not have escaped beyond the Hubble horizon by the time the observer is
> aware of the results of the experiment. So, whether or not the photons
> will eventually no longer be accessible, cannot be relevant.
>


Once the photons escape from the immediate environs of the experiment, they
are not recoverable. Try shining a torch at night to illuminate a tree. Now
try to stop the illumination already present. You can stop future
illumination by covering the torch, or switching it off. But once the tree
is illuminated it is not reversible. The expansion of the universe, and the
existence of the Hubble horizon, just makes the irreversibility more
obvious.

Bruce

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Re: Why are laws of physics stable?

On 10-07-2021 21:41, 'Brent Meeker' via Everything List wrote:

On 7/10/2021 1:19 AM, Bruno Marchal wrote:
So, in general we can conclude by generalizing this to any large
number of particles that even with what we consider to be permanent
records, you don't get rid of the theoretical possibility of
interference between the sectors where those records are different.

We can if the universe is expanding faster than light beyond the Hubble
radius.

The expansion of the universe is irrelevant. The information needed to
see the interference pattern continues to exist outside the horizon when
it isn't accessible to us anymore. And this is irrelevant for the
discussions about observations in quantum mechanics. If an observer
performs a measurement and the claim is that this is a unitary process
with the observer evolving into a superposition, while the objection
against this claim is that infrared photons are escaping and will
eventually move beyond the Hubble volume, then these photons will still
not have escaped beyond the Hubble horizon by the time the observer is
aware of the results of the experiment. So, whether or not the photons
will eventually no longer be accessible, cannot be relevant.

Saibal

Brent

Re: Why are laws of physics stable?

2021-07-10 Thread 'Brent Meeker' via Everything List

On 7/10/2021 12:20 PM, Tomas Pales wrote:

On Saturday, July 10, 2021 at 2:49:07 PM UTC+2 johnk...@gmail.com wrote:

On Mon, Jul 5, 2021 at 4:19 PM Tomas Pales wrote:

/>You are conscious of certain parts of your brain/

I've never actually seen it

Your eyes receive visual signals from outside your head, not from the
inside, so you don't see your own brain.

soif I hadn't read about human anatomy in books I wouldn't even
know that I had a brain. How can I be conscious of something that
I don't even know exists?

Neuroscience says that we are not directly conscious of the external
world but we are directly conscious of its neural representations in
our brain. I would say that's because we/are/ those representations.

I think this kind of talk puts far too much on consciousness. Conscious
thoughts seem to pop into my head with no antecedents, yet they relate
to past and distant things in my experience. The Poincare' effect shows
that even the most abstract thought is largely unconscious.

Brent

So you are conscious of parts of your brain but not of those
properties of the brain that look like gray matter when they are
looked at through the eyes. Instead, the properties of the brain you
are conscious of look like a red tomato, taste like chocolate or sound
like music.

You may wonder how can a piece of gray matter look like a red tomato?
I think it's because only those properties of the brain matter that
are perceptible by the eyes look gray. The brain matter has many other
properties that cannot be seen but some of them may feel like when you
are looking at a red tomato.

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Re: Why are laws of physics stable?

2021-07-10 Thread 'Brent Meeker' via Everything List

On 7/10/2021 2:00 AM, Bruno Marchal wrote:

On 6 Jul 2021, at 22:07, 'Brent Meeker' via Everything List
> wrote:

On 7/6/2021 10:34 AM, Jason Resch wrote:

On Tue, Jul 6, 2021 at 12:27 PM 'Brent Meeker' via Everything List
> wrote:

And you're never going to find a being that behaves
intelligently based on information that can be quantum erased.

You need only a quantum computer with enough qubits.

Can you prove that? How does this quantum intelligence ever arrive
at a definite decision?

With a quantum brain, you can hack all credit cards, by running shor
algorithm in your head for example.

You argument seems to negate the possibility of quantum speeding. You
would be right if P = NP, or something…

No. As I recall even Shor's algorithm has a probabilistic step so that
the answer is only correct with high probability, not certainty.
Quantum speedup is possible, just not quantum definiteness. One you or
your QC decides to act that act cannot be a superposition of actions.
And the decoherence is not just FAPP, it is inherent in the loss of
information across the Hubble boundary.

Brent

Bruno

Brent

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Re: Why are laws of physics stable?

2021-07-10 Thread 'Brent Meeker' via Everything List

On 7/10/2021 1:48 AM, Bruno Marchal wrote:

On 6 Jul 2021, at 16:45, Tomas Pales > wrote:

On Tuesday, July 6, 2021 at 10:28:11 AM UTC+2 Bruno Marchal wrote:

On 3 Jul 2021, at 14:13, Tomas Pales wrote:
Can't there be a machine that computes gravitational interaction
with gravitational constant 6.674 x 10 to the -11 up until some
time t and then continues the computation with gravitational
constant 5 x 10 to the -11, or just halts? That would be an
instability or cessation of gravitational law.

Yes, and that exists, but such world will have a very low
probability to be accessed by any observer, due to the fact that,
below our mechanist substitution level, all such theories intervene.

How low is this probability? Is it maybe as low as the probability
that my whole body quantum-tunnels through a wall?

Yes.

What does it mean that "below our mechanist substitution level, all
such theories intervene”?

If your consciousness does not require some details: like the position
of an electron of some atom in some neurotransmitter (say), then it
will be associate as much with your brain and that election here, and
with you brain and that electron there, and if you don’t measure the
position of that electron, the two histories can interfere statistically.

What would it mean to measure the state of your own brain?

Brent

An electronic orbital is such a map: it tells you where you can find
the electron in your most probable computational histories.

Bruno

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Re: Why are laws of physics stable?

2021-07-10 Thread 'Brent Meeker' via Everything List

On 7/10/2021 1:28 AM, Bruno Marchal wrote:

On 6 Jul 2021, at 12:55, John Clark > wrote:

On Sun, Jul 4, 2021 at 10:10 PM 'Brent Meeker' via Everything List
> wrote:

>> It's easy to determine that the quantum computer is
intelligent butas for consciousness, how did you determine
that it was not conscious?For that matter how did you
determine that I am conscious? But let's get out of the
consciousness quagmire for a moment so I can ask you a
question, leaving behind the interpretation of the experiment
concentrating only on its results, if it was actually
performed as described do you think interference bands would
be on that photographic plate or would there be no such
bands? I would bet money the bands would be there on that
plate even though there's no longer any which way information
remaining. So, what would you put your money on, bands or no
bands?

> /I would guess the interference bands would be present exactly
because, ex hypothesi, the which-way information was quantum erased.
/

So an intelligent and presumably consciousbeingonce existed that knew
which slot all the electrons went through, but those interference
bands still showed up anyway. Don't you find that a little strange?
If Many Worlds is wrong and that being didn't exist in another world,
then where did it exist?

It, or rather it's knowledge, existed in the past, before being erased.

Brent

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Re: Why are laws of physics stable?

2021-07-10 Thread 'Brent Meeker' via Everything List





On 7/10/2021 1:19 AM, Bruno Marchal wrote:

So, in general we can conclude by generalizing this to any large number of 
particles that even with what we consider to be permanent records, you don't 
get rid of the theoretical possibility of interference between the sectors 
where those records are different.


We can if the universe is expanding faster than light beyond the Hubble 
radius.


Brent

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Re: Why are laws of physics stable?

2021-07-10 Thread 'Brent Meeker' via Everything List

On 7/10/2021 1:06 AM, Bruno Marchal wrote:

On 5 Jul 2021, at 21:01, 'Brent Meeker' via Everything List
> wrote:

On 7/5/2021 7:41 AM, John Clark wrote:
On Mon, Jul 5, 2021 at 9:44 AM Tomas Pales > wrote:

>> "Brain" is a noun, "consciousness" is not, that's why you
can't measure consciousness by the pound or by the cubic inch.

/> In English language it is used as a noun. Check out a
dictionary:/

*consciousness* noun

I know, that's what my fourth grade teacher told me too, but I
long-ago realized that neither she nor the lexicographerswho wrote
that big thick book are the fonts of all wisdom.

>> Intelligence is what a brain does not what a brain is, and
because Darwinian Evolution is almost certainly correct,
consciousness must be an inevitable byproduct of
intelligence, therefore "consciousness" is not a noun, it's
a word that describes what a noun (in this case the brain)
does, in other words consciousness is an adjective.

/> You mean a verb then, no? /

I think adjective fits the bill a littlebetter,I think Tomas Palesis
the way atoms behave when they are arranged in a Tomaspalesian way.

> /consciousness is a spatiotemporal object./

I disagree, I think asking where my consciousness is located would
be like asking where the number 11 or the color yellow or "fast" is
located. If my brain is in Paris and I'm looking at a TV football
game from Detroit and I'm listening to a friend in Australia on my
telephone and I'm thinking about The Great Wall of China would it
make sense to say my consciousness is really located inside a box
made of bone mounted on my shoulders when I have no conscious
experience of being in a bone box on my shoulders? I don't think so.

Yet a sharp blow to that bone box would eliminate your conscious
experience at least temporarily.

Only from the point of view of some conscious subject. From the point
of view of the person associated to the brain in the box, that does
not make sense, as it is associated to infinitely many truing
universal relation.

That's incorrect. I've been knocked unconscious and when I regained
consciousness (it was on a few seconds) I realized the gap my conscious
experience.

The body is only a map on infinitely many histories. That can be
proved both with QM-without-collapse, or in any non trivial
combinatory algebra (like a model of arithmetic).

So there's something there that is essential to your consciousness.

What is “essential” are the infinitely many computations.

Since the 1930s we know that all computations are realised in any
model (in the logician sense) of arithmetic, or of combinatory logic
(Kxy = x, Sxyz = xz(yz)).

But not in any brain...they are only finite.

I know that this contradict 1492 years of materialist brainwashing,
but “appearance of matter” are explained in arithmetic, and get
contradictory when associated or singularised through any
supplementary axioms, even the induction axioms used to define what an
observer can be.

You assume some ontological commitment inconsistent with Mechanism here.

You assume an ontological commitment to Church-Turing infinite
computations.

Brent

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Re: Why are laws of physics stable?

2021-07-10 Thread Tomas Pales

On Saturday, July 10, 2021 at 2:49:07 PM UTC+2 johnk...@gmail.com wrote:

> On Mon, Jul 5, 2021 at 4:19 PM Tomas Pales  wrote:
>
> *>You are conscious of certain parts of your brain*
>>
>
> I've never actually seen it 
>

Your eyes receive visual signals from outside your head, not from the 
inside, so you don't see your own brain.

> so if I hadn't read about human anatomy in books I wouldn't even know that 
> I had a brain. How can I be conscious of something that I don't even know 
> exists?
>

Neuroscience says that we are not directly conscious of the external world 
but we are directly conscious of its neural representations in our brain. I 
would say that's because we* are* those representations. So you are 
conscious of parts of your brain but not of those properties of the brain 
that look like gray matter when they are looked at through the eyes. 
Instead, the properties of the brain you are conscious of look like a red 
tomato, taste like chocolate or sound like music.

You may wonder how can a piece of gray matter look like a red tomato? I 
think it's because only those properties of the brain matter that are 
perceptible by the eyes look gray. The brain matter has many other 
properties that cannot be seen but some of them may feel like when you are 
looking at a red tomato.

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Re: Why are laws of physics stable?


On 10-07-2021 07:58, Bruce Kellett wrote:

On Sat, Jul 10, 2021 at 10:21 AM smitra  wrote:


On 08-07-2021 01:51, Bruce Kellett wrote:


Do you dispute that that is what the paper by Hornberger et al.

says?


Bruce


I don't dispute these results. The buckyballs are coming from a
thermal
reservoir at some finite temperature. We can avoid working with
mixed
states by simply considering the interference pattern for each pure
state separately and then summing over the probability distribution
over
the pure states. But for this discussion we want to focus on what
the
interference pattern will be if all the buckyballs are in the same
exited state. If we put the entire system ina finite volume then we
have
a countable set of allowed k-values for the photon momenta. We then
have
a set of allowed states for the photons defined by the allowed
momenta
and a polarization. We can then label these photon states using a
number
and then specify an arbitrary state for the photons by specifying
how
many photons we have in each state, and these numbers can then be
equal
to zero. If they are all zero then no photons are present.

If only the right slit is open, the state of the buckyball and the
photons just before the screen is hit can be denoted as:

|Right> = sum over n1, n2,n3,...|R(n1,n2,n3,...)>|n1,n2,n3,..>

where |R(n1,n2,...)> denotes the quantum state of the buckyball if
it
emits n1 photons in state 1, n2 photons in state 2 etc. The state of
the
photons is then denoted as |n1,n2,n3,..>

If only the left slit is open, the state of the buckyball and the
photons just before the screen is hit can be denoted as:

|Left> = sum over n1, n2,n3,...|L(n1,n2,n3,...)>|n1,n2,n3,..>

where |R(n1,n2,...)> denotes the quantum state of the buckyball.
Here we
note that the state of the photons will pick up a phase factor
relative
to the case of only the right slit being open, but we can then
absorb
this phase factor in |L(n1,n2,n3,...)>.

With both slits open, we'll then have a state of the form:

|psi> = 1/sqrt(2) [|Right> + |Left>]

The inner product of |psi> with some position eigenstate |x>,
 is
then a state vector for the photon states, the squared norm of that
state vector is the probability of finding the buckyball at position
x,
because this is the sum over the probabilities for photons over all
possible photon states. So, the probability is then:

P(x) = ^2 = 1/2 [^2 + ^2] +
Re[]

We can then evaluate the interference term as follows:

Re[] = Re sum over n1, n2,n3,...m1,m2,m3


*


Using that  = 0 unless m_j = n_j for
all
j, in which case this inner product equals 1, we then have:

Re[] = Re sum over n1, n2,n3
* =

Re[*] +
Re[*] + .  (1)

As explained above when there are photons present then we've
absorbed
the phase factor due to translation of the photon states in the
states
|L(n1,n2,n3...)>. For each wave vector k there is a factor for each
photon with that wavevector of exp(i k dot r) where r is the
position of
the left slit w.r.t. the right slit. So, the total phase factor will
be:

Product over j of exp(i kj nj dot r)

In the experiment there is then an additional summation over the
pure
states of the buckyballs. If the temperature is low then the
summation
will consist of states for which |R(0,0,0,..> and |L(0,0,0,..> are
the
dominant terms, as most of the time no photons will be emitted. At
higher temperatures the typical states there will be contributions
from
different numbers pf photons, so the interference pattern will be a
sum
of many different terms in (1) with comparable norms, they come with

different phase factors due to the different numbers of photons with

different momenta. So, the interference pattern will be washed out.


This analysis contradicts what you said in your first analysis. In the
first analysis, you claimed that no interference would be seen if the
IR photons were not detected.


In that case I considered a simple model where one photon would be 
emitted such that the state of that photon emitted from the ball moving 
through one slit would be orthogonal to that of the state of the photon 
emitted by the ball moving through the opther slit.



You seem to have dropped this notion in

the above.


In the derivation above I work in the momentum basis for the photons. 
Obviously, exp(i k dot r) and exp(i k dot (r+u)) are not orthogonal, but 
then one needs to consider a superposition of such momentum eigenstates 
as I do above.


 You now say that there is interference when no photons are

present, but this is washed out when there are different numbers of
photons.



Yes, the above derivation captures more of the details.


I think you are making the same basic mistake that you made when we
previously discussed  two-slit interference: You are analysing the two
slit case as a sum over single slit patterns, and assuming the
emission of independent photons from the ball through each slit. The
trouble is that there are not two balls, one for each slit. The same

Re: Why are laws of physics stable?

2021-07-10 Thread John Clark

On Mon, Jul 5, 2021 at 4:19 PM Tomas Pales  wrote:

>> How can my consciousness be located in a place that I am not conscious
>> of?
>>
>
> *>You are conscious of certain parts of your brain*
>

I've never actually seen it so if I hadn't read about human anatomy in books I
wouldn't even know that I had a brain. How can I be conscious of something
that I don't even know exists?  And for thousands of years people thought
the heart was the earthly seat of the soul and consciousness, and they
thought the brain was just some goop inside the head of no more
philosophical significance than snot.

*> We can't rule out that your consciousness is not located in your brain*
>

But we don't even need to perform experiments to rule out the brain being
the location of consciousness because it is a logical absurdity to say you
are conscious of something you were not conscious of.  I certainly am not
conscious of being imprisoned in a box made of bone, and I very much doubt
anybody else is either.

John K ClarkSee what's on my new list at  Extropolis

bqm1

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Re: Why are laws of physics stable?

2021-07-10 Thread John Clark

On Sat, Jul 10, 2021 at 3:52 AM Bruno Marchal  wrote:

>> "Brain" is a noun, "consciousness" is not
>
>

*> I disagree with this *

An embalmed brain rotting in a grave is a noun. Do you therefore think it's
conscious? I don't because it's not doing anything that 3 pounds of rotting
hamburger isn't doing, and neither of the two are behaving intelligently.

>> therefore "consciousness" is not a noun, it's a word that describes what
>> a noun (in this case the brain) does, in other words consciousness is an
>> adject
>
>
> *> This is logically inconsistent with Descartes Mechanism. *

I have no idea what  "*Descartes Mechanism*" is, and after listening to you
all these years I am quite certain you can't give a coherent explanation
for it either, but whatever it means if it is logically inconsistent with
what I said then "*Descartes Mechanism*" is wrong.
*> Without mechanism, it is consistent, but still problematic with Occam
razor *

As I said before,  Occam razor is about economy of assumptions not economy
of results.

John K ClarkSee what's on my new list at  Extropolis

0o6

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Re: Why are laws of physics stable?

2021-07-10 Thread Jason Resch

On Sat, Jul 10, 2021, 1:58 AM Bruce Kellett  wrote:

> On Sat, Jul 10, 2021 at 10:21 AM smitra  wrote:
>
>> On 08-07-2021 01:51, Bruce Kellett wrote:
>> >
>> > Do you dispute that that is what the paper by Hornberger et al. says?
>> >
>> > Bruce
>>
>> I don't dispute these results. The buckyballs are coming from a thermal
>> reservoir at some finite temperature. We can avoid working with mixed
>> states by simply considering the interference pattern for each pure
>> state separately and then summing over the probability distribution over
>> the pure states. But for this discussion we want to focus on what the
>> interference pattern will be if all the buckyballs are in the same
>> exited state. If we put the entire system ina finite volume then we have
>> a countable set of allowed k-values for the photon momenta. We then have
>> a set of allowed states for the photons defined by the allowed momenta
>> and a polarization. We can then label these photon states using a number
>> and then specify an arbitrary state for the photons by specifying how
>> many photons we have in each state, and these numbers can then be equal
>> to zero. If they are all zero then no photons are present.
>>
>> If only the right slit is open, the state of the buckyball and the
>> photons just before the screen is hit can be denoted as:
>>
>> |Right> = sum over n1, n2,n3,...|R(n1,n2,n3,...)>|n1,n2,n3,..>
>>
>> where |R(n1,n2,...)> denotes the quantum state of the buckyball if it
>> emits n1 photons in state 1, n2 photons in state 2 etc. The state of the
>> photons is then denoted as |n1,n2,n3,..>
>>
>>
>> If only the left slit is open, the state of the buckyball and the
>> photons just before the screen is hit can be denoted as:
>>
>> |Left> = sum over n1, n2,n3,...|L(n1,n2,n3,...)>|n1,n2,n3,..>
>>
>> where |R(n1,n2,...)> denotes the quantum state of the buckyball. Here we
>> note that the state of the photons will pick up a phase factor relative
>> to the case of only the right slit being open, but we can then absorb
>> this phase factor in |L(n1,n2,n3,...)>.
>>
>> With both slits open, we'll then have a state of the form:
>>
>> |psi> = 1/sqrt(2) [|Right> + |Left>]
>>
>> The inner product of |psi> with some position eigenstate |x>,  is
>> then a state vector for the photon states, the squared norm of that
>> state vector is the probability of finding the buckyball at position x,
>> because this is the sum over the probabilities for photons over all
>> possible photon states. So, the probability is then:
>>
>> P(x) = ^2 = 1/2 [^2 + ^2] +
>> Re[]
>>
>>
>> We can then evaluate the interference term as follows:
>>
>> Re[] = Re sum over n1, n2,n3,...m1,m2,m3
>> *
>>
>> Using that  = 0 unless m_j = n_j for all
>> j, in which case this inner product equals 1, we then have:
>>
>> Re[] = Re sum over n1, n2,n3
>> * =
>>
>>
>> Re[*] +
>> Re[*] + .  (1)
>>
>> As explained above when there are photons present then we've absorbed
>> the phase factor due to translation of the photon states in the states
>> |L(n1,n2,n3...)>. For each wave vector k there is a factor for each
>> photon with that wavevector of exp(i k dot r) where r is the position of
>> the left slit w.r.t. the right slit. So, the total phase factor will be:
>>
>> Product over j of exp(i kj nj dot r)
>>
>> In the experiment there is then an additional summation over the pure
>> states of the buckyballs. If the temperature is low then the summation
>> will consist of states for which |R(0,0,0,..> and |L(0,0,0,..> are the
>> dominant terms, as most of the time no photons will be emitted. At
>> higher temperatures the typical states there will be contributions from
>> different numbers pf photons, so the interference pattern will be a sum
>> of many different terms in (1) with comparable norms, they come with
>> different phase factors due to the different numbers of photons with
>> different momenta. So, the interference pattern will be washed out.
>>
>
>
> This analysis contradicts what you said in your first analysis. In the
> first analysis, you claimed that no interference would be seen if the IR
> photons were not detected. You seem to have dropped this notion in the
> above. You now say that there is interference when no photons are present,
> but this is washed out when there are different numbers of photons.
>
> I think you are making the same basic mistake that you made when we
> previously discussed  two-slit interference: You are analysing the two slit
> case as a sum over single slit patterns, and assuming the emission of
> independent photons from the ball through each slit. The trouble is that
> there are not two balls, one for each slit. The same ball goes through both
> slits and there is only one ball emitting (or not emitting) photons.
>
> Neither of your analyses actually explain the observed behaviour.
>

This conversation brought to mind Feynman's question: what happens in the
limit of going from 2 slits to infinite slits, such that

Re: Why are laws of physics stable?



> On 7 Jul 2021, at 09:34, smitra  wrote:
> 
> On 06-07-2021 22:29, 'Brent Meeker' via Everything List wrote:
>> On 7/6/2021 12:49 PM, smitra wrote:
>>> On 06-07-2021 19:34, Jason Resch wrote:
 On Tue, Jul 6, 2021 at 12:27 PM 'Brent Meeker' via Everything List
  wrote:
 And you're never going to find a being that behaves intelligently
 based on information that can be quantum erased.
 You need only a quantum computer with enough qubits.
 Jason
>>> Indeed, the critics have to show how the laws of physics imply that 
>>> decoherence cannot be limited to the extent necessary to run a good enough 
>>> quantum computer simulation of an entire brain for this to work. And one 
>>> has lots of elbowroom available for the thought experiment. Practical issue 
>>> that would make this unfeasible for us to do play no role at all, but real 
>>> physical limits would be valid objections. The amount of available 
>>> resources that can be used physically is at least a large fraction of all 
>>> the materials that are present in our galaxy. One can build Dyson spheres 
>>> around a far fraction of all stars in the galaxy, the available time is at 
>>> least of the order of tens of billions of years. The simulation does not 
>>> have to run in real time, each simulated second can take a billion years, 
>>> which may be necessary to perform enough quantum error correction to make 
>>> this work.
>>> If it can be shown that under more generous conditions this is not 
>>> feasible, so large scale quantum computing is not going to work even with 
>>> most of the resources in the observable universe, and that a large scale 
>>> computation needed for the thought experiment cannot be finished before the 
>>> end of the universe, then the critics have a point. But even then it's only 
>>> a hint of a problem, because the objection would only be consistent with 
>>> the unproven hypothesis that unitary time evolution breaks down when a 
>>> large enough number of degrees of freedom get entangled with a quantum 
>>> system.
>>> Saibal
>> Why are you worrying about enormous quantum computers?  A quantum
>> computer should have much more computational power than a classical
>> computer and we already know of an intelligent classical computer fits
>> in a little more than a liter.  The problem isn't computational power,
>> it's reaching definite answer.  Quantum computers in general provide a
>> readout by decoherence, and then it's no longer erasable.
>> Brent
> 
> 
> There can never be a definite answer as QM is unitary and decoherence is 
> never complete. If you assume that the real world is fundamentally different 
> from a virtual world simulated by a quantum computer, no matter how large 
> that quantum simulation, then you are assuming that the real world violates 
> QM in an essential way.

Yes. That is why Bohr has to distinguish the micro-qunatum reality, and the 
classical macro realm. To believe in a collapse is the same as to believe that 
the SWE is wrong, and can’t be applied to the observer. Bohr was explicitly 
dualist. He even believed that the SWE would be wrong at smaller scale.

The SWE is the same as the MANY-WORLDS, that is why the pioneer feels necessary 
to add the collapse postulate, but Everett showed clearly that this was neither 
necessary, nor even consistent with the idea that the observer obeys to the SWE.

Now, we know that Bruce and John disbelieves in the self-indeterminacy in 
self-multiplication experience, which is inconsistent at a much more basic 
level, and eventually is the usual confusion between []p and []p & p, i.e. the 
confusion between first person description and third person description. That 
is of course vital for understanding SWE without collapse.

Bruno



> 
> Saibal
> 
> 
> 
> 
> 
> 
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Re: Why are laws of physics stable?


> On 7 Jul 2021, at 03:21, Bruce Kellett  wrote:
> 
> On Wed, Jul 7, 2021 at 11:13 AM Jason Resch  > wrote:
> On Tue, Jul 6, 2021, 2:03 PM 'Brent Meeker' via Everything List 
> mailto:everything-list@googlegroups.com>> 
> wrote:
> Then I guess I don't understand this part:
> 
> Run it together with Shors algorithm and have "each AI" read a definite 
> random number from 0 to 2^n where n is the number of qubits needed to 
> represent the semiprime being factored. Then have the AI copy that number to 
> another register to prove it went through the AI's mind.
> 
> What does it mean to "read a definite random number"
> 
> F(x) is a quantum algorithm (a combination AI + Shor's algorithm) which takes 
> an input x where x is a set of N qubits, with each qubit initialized to a 
> superposition of 1 and 0.
> 
> Since the qubits are in a superposition representing 2^N states, the quantum 
> algorithm likewise becomes a superposition of 2^N uniquely processed values. 
> Each one can be viewed as a unique evaluation of F(i) where i is each of the 
> possible N-bit bit strings.
> 
> Since F() includes a conscious AI evaluating the input value, and since it 
> exists in a superposition, then the evaluation on a quantum computer 
> corresponds to 2^N independent conscious states.
> 
> 
> and what does that have to do with recording which slit a photon went thru?
> 
> 
> It's an alternate example of Deutsch's experiment which shows that 
> consciousness doesn't cause collapse, assuming adding a conscious AI to 
> Shor's algorithm doesn't somehow break the algorithm. If you can still factor 
> numbers with the AI added to the circuit, then consciousness doesn't cause 
> collapse, and we can see QM directly leads to many "split" observers.
> 
> 
> No one now  believes that consciousness has anything to do with collapse. For 
> example, in fGRW, the collapse is caused by independent stochastic 'flashes' 
> that have no relevance to consciousness. In Bohm's theory, there never is any 
> collapse because there is never any mystic 'superposition’.


Bohm’s theory keep the wave, and the superposition. It just that the particles 
stay in only one term of that superposition, yet remained influenced (even at a 
FTL speed) by what happens in the superposition. It is the many-worlds, + a 
complex supplementary potential to make infinitely many people into zombies, so 
that consciousness is singularised in one branch only.
It is heavy metaphysical assumption to satisfy personal coquetry...



> In Penroses gravitational induced collapse, the collapse is due to changes in 
> the spacetime metric -- again, independent of consciousness.

Von Neumann and Wigner were the early advocate of the “consciousness makes the 
collapse” idea. 
I consider that it has been refuted correctly by Abner Shimony (and many 
others).



> 
>  So Deutsch's thought experiment is about nothing at all, and proves nothing 
> at all.


In that experiment, the Many)worlds predict that the observer can remember 
having seen as definite result, without remembering each one, and that, if 
done, would add confirmation to the MW, or better, many histories or many 
computations, like arithmetic (+ mechanism) predicted.

Bruno




> 
> Bruce
> 
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>  
> .

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Re: Why are laws of physics stable?


> On 6 Jul 2021, at 22:07, 'Brent Meeker' via Everything List 
>  wrote:
> 
> 
> On 7/6/2021 10:34 AM, Jason Resch wrote:
>> 
>> 
>> On Tue, Jul 6, 2021 at 12:27 PM 'Brent Meeker' via Everything List 
>> mailto:everything-list@googlegroups.com>> 
>> wrote:
>> And you're never going to find a being that behaves intelligently based on 
>> information that can be quantum erased.
>> 
>> You need only a quantum computer with enough qubits.
> Can you prove that?  How does this quantum intelligence ever arrive at a 
> definite decision?
> 
> 

With a quantum brain, you can hack all credit cards, by running shor algorithm 
in your head for example.

You argument seems to negate the possibility of quantum speeding. You would be 
right if P = NP, or something…

Bruno


> Brent
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Re: Why are laws of physics stable?


> On 6 Jul 2021, at 19:22, spudboy100 via Everything List 
>  wrote:
> 
> At an open AA meeting I once attended,  some atheist-leaning recovering 
> alcoholics once joked that GOD meant Group Of Drunks. 

LOL.

I read that today it means mainly:  Gold-Oil-Drugs…

Group of Drunks is closer to the original meaning. Once theology has been 
stolen from science, the first interdiction was the (otherwise common) use of 
entheogen, like alcohol, which still remain present in the Church...

:)

Bruno



> 
> 
> -Original Message-
> From: John Clark 
> To: 'Brent Meeker' via Everything List 
> Sent: Tue, Jul 6, 2021 10:56 am
> Subject: Re: Why are laws of physics stable?
> 
> 
> On Tue, Jul 6, 2021 at 4:49 AM Bruno Marchal  <mailto:marc...@ulb.ac.be>> wrote:
> 
> > Many people continue to believe in the God “Matter”
> 
> Yeah, many people are like that, people such as yourself who long ago have 
> given up on the idea of God because it is ridiculous but love the way the 
> English word  "G-O-D" sounds so much that they refuse to abandon it.
> 
> John K ClarkSee what's on my new list at  Extropolis 
> <https://groups.google.com/g/extropolis>
> stty
>> lts9
> 
> 
>  
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Re: Why are laws of physics stable?


> On 6 Jul 2021, at 16:45, Tomas Pales  wrote:
> 
> 
> On Tuesday, July 6, 2021 at 10:28:11 AM UTC+2 Bruno Marchal wrote:
> 
>> On 3 Jul 2021, at 14:13, Tomas Pales > > wrote:
>> Can't there be a machine that computes gravitational interaction with 
>> gravitational constant 6.674 x 10 to the -11 up until some time t and then 
>> continues the computation with gravitational constant 5 x 10 to the -11, or 
>> just halts? That would be an instability or cessation of gravitational law.
> 
> Yes, and that exists, but such world will have a very low probability to be 
> accessed by any observer, due to the fact that, below our mechanist 
> substitution level, all such theories intervene.
> 
> How low is this probability? Is it maybe as low as the probability that my 
> whole body quantum-tunnels through a wall?

Yes.



> 
> What does it mean that "below our mechanist substitution level, all such 
> theories intervene”?


If your consciousness does not require some details: like the position of an 
electron of some atom in some neurotransmitter (say), then it will be associate 
as much with your brain and that election here, and with you brain and that 
electron there, and if you don’t measure the position of that electron, the two 
histories can interfere statistically.

An electronic orbital is such a map: it tells you where you can find the 
electron in your most probable computational histories.

Bruno 




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Re: Why are laws of physics stable?

> On 6 Jul 2021, at 16:56, John Clark  wrote:
> 
> 
> On Tue, Jul 6, 2021 at 4:49 AM Bruno Marchal  > wrote:
> 
> > Many people continue to believe in the God “Matter”
> 
> Yeah, many people are like that, people such as yourself who long ago have 
> given up on the idea of God because it is ridiculous but love the way the 
> English word  "G-O-D" sounds so much that they refuse to abandon it.

I just use Plato’s definition. It is Aristotle who invented a supplementary 
spurious God along a “creation”. 

Replace “GOD” by any ontological commitment in some fundamental reality. GOD is 
defined by whatever you assume to explain all the rest.

You are the one believing in a mysterious, never defined, primitive 
materiality, and you use it to select some computations in arithmetic to make 
them more real than all of them in arithmetic, but that does not just introduce 
an infinity of zombies in arithmetic, I also confer something not Turing 
emulable playing a role in the brain, forcing the abandon of Mechanism. 

Bruno

> 
> John K ClarkSee what's on my new list at  Extropolis 
> 
> stty
>> lts9
> 
> 
>  
> 
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Re: Why are laws of physics stable?

> On 6 Jul 2021, at 16:39, Tomas Pales  wrote:
> 
> 
> On Tuesday, July 6, 2021 at 10:12:49 AM UTC+2 Bruno Marchal wrote:
> 
> The physical laws are stable because they have an arithmetical origin in the 
> “head” of any universal+ machine (those which have the theology G*)
> 
> What do you mean by "head”?

Here, I meant the mind of the universal machine, like in the expression “it is 
all in your head”, which means “it is a product of your imagination”.

With Indexical Digital Mechanism (“yes doctor (for a digital body transplant” + 
the Church-Turing thesis), the physical reality is a statistics on all relative 
computations going through “my state”. There is an infinity of such 
computations, in all models of any Turing complete theory. 

You might read my grand public presentation in Amsterdam in 2004: 
B. Marchal. The Origin of Physical Laws and Sensations. In 4th International 
System Administration and Network Engineering Conference, SANE 2004, Amsterdam, 
2004.
http://iridia.ulb.ac.be/~marchal/publications/SANE2004MARCHALAbstract.html 

Bruno

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Re: Why are laws of physics stable?


> On 6 Jul 2021, at 15:53, Jason Resch  wrote:
> 
> 
> 
> On Mon, Jul 5, 2021, 4:19 PM Tomas Pales  > wrote:
> 
> On Monday, July 5, 2021 at 8:03:46 PM UTC+2 johnk...@gmail.com 
>  wrote:
> 
> How can my consciousness be located in a place that I am not conscious of?
> 
> You are conscious of certain parts of your brain (presumably those that have 
> high organized complexity of a certain kind), many of which are 
> representations of external objects. If there are no representations of your 
> skull it is probably because they have not been useful in evolution. It is 
> more useful to be conscious of what is going on around your body than of the 
> interior of your skull.
> 
> We can't rule out that your consciousness is not located in your brain and 
> your brain just serves as a kind of preliminary processor of sensory data 
> that somehow sends the preprocessed data to some other object (soul?) where 
> your consciousness is located but there's not much evidence for that.
> 
> 
> This conversation reminds me of Daniel Dennett's short essay "Where am I?":
> 
> https://www.lehigh.edu/~mhb0/Dennett-WhereAmI.pdf 
> 


That is very good. It is the place where Dennett get very close, but miss, the 
first person indeterminacy, note.

Bruno



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Re: Why are laws of physics stable?


> On 6 Jul 2021, at 12:55, John Clark  wrote:
> 
> 
> On Sun, Jul 4, 2021 at 10:10 PM 'Brent Meeker' via Everything List 
> mailto:everything-list@googlegroups.com>> 
> wrote:
> 
> >> It's easy to determine that the quantum computer is intelligent but as for 
> >> consciousness, how did you determine that it was not conscious? For that 
> >> matter how did you determine that I am conscious? But let's get out of the 
> >> consciousness quagmire for a moment so I can ask you a question, leaving 
> >> behind the interpretation of the experiment concentrating only on its 
> >> results, if it was actually performed as described do you think 
> >> interference bands would be on that photographic plate or would there be 
> >> no such bands? I would bet money the bands would be there on that plate 
> >> even though there's no longer any which way information remaining. So, 
> >> what would you put your money on, bands or no bands? 
> > I would guess the interference bands would be present exactly because, ex 
> > hypothesi, the which-way information was quantum erased.
> 
> So an intelligent and presumably conscious being once existed that knew which 
> slot all the electrons went through, but those interference bands still 
> showed up anyway. Don't you find that a little strange? If Many Worlds is 
> wrong and that being didn't exist in another world, then where did it exist? 
> 
> >> If interference bands are on that photographic plate then either Many 
> >> Worlds is correct or a rock is just as likely to be conscious as one of 
> >> your fellow human beings because intelligent behavior would tell you 
> >> nothing about consciousness. But if there are no bands I would immediately 
> >> concede and say Many Worlds must be wrong. What outcome would make you 
> >> concede? 
> > Concede what? 
> 
> What experimental evidence would make you concede that your theory that Many 
> Worlds must be wrong, is wrong. Or is your theory by its very nature 
> unprovable? My theory that Many Worlds is less wrong than other quantum 
> interpretations at least has the virtue of being capable of being proven 
> wrong. Let me put the question to you this way, what conclusion would you 
> draw if you saw interference bands on that photographic plate, and what 
> conclusion would you draw if you DID NOT SEE interference bands on that 
> photographic plate?
> > You're the one that cast the hypothetical in terms of consciousness.
> 
> I only said that because some (but not me) claim Quantum Mechanics has 
> something to do with consciousness, so if you want to test that claim 
> experimentally the first thing you're going to need is something people can 
> agree on that is conscious; and I don't think you're ever going to find 
> anything better for that "something" than a being that behaves intelligently.


I agree that using the quantum to explain consciousness does not work, despite 
the initial motivation brought by the wave collapse is not entirely 
meaningless, but this has failed (cf Abner Shimony). 

Without collapse, physics confirms the many-histories or many-dreams 
interpretation of arithmetic on which all universal numbers converge, again in 
arithmetic.

Consciousness is given by knowledge, which is given by the modes of 
self-reference imposed by incompleteness, (those with “& p”) and the fact that 
universal+ (Löbian) machine are aware of it.

P
[]p
[]p & p [HERE]
[]p & <>t
[]p & <>t & p. [HERE]

[]p & p gives knowledge, and []p & <>t & p gives a form of non transitive 
(immediate) physical consciousness.

The main facts are that G* (the theology) proves []p <-> []p & p, and that G, 
the machine itself, does not, making them obeying quite different logics and 
mathematics.

Bruno




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> 
> rroo
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Re: Why are laws of physics stable?



> On 6 Jul 2021, at 09:21, smitra  wrote:
> 
> On 05-07-2021 12:18, Bruce Kellett wrote:
>> On Mon, Jul 5, 2021 at 7:39 PM smitra  wrote:
>>> On 05-07-2021 09:00, Bruce Kellett wrote:
 On Mon, Jul 5, 2021 at 2:23 PM smitra  wrote:
 I don't think this is actually done in the experiment. What is
 observed is the presence or absence of the interference pattern on
>>> the
 screen where the balls hit. The photons are not detected. But if,
>>> in
 principle, they are of suitable wavelength to resolve the slit
 difference, then the interference pattern vanishes. The experiment
>>> is
 convincing in that they start wil cold buckyballs which show a
>>> clear
 interference pattern. They then gradually heat the balls so that
>>> the
 typical wavelength of the photons decreases. This gradually washes
>>> out
 the interference pattern. (Because at lower temperatures, the
 wavelength distribution of the IR photons is such that a few of
>>> them
 have shorter wavelengths.) As the temperature is increased so that
 most IR photons have short enough wavelengths, the interference
 pattern disappears completely. The paper by Hornberger et al. is
>>> at
 arXiv:quant-ph/0412003v2
>>> This is then what I said previously, what you denied, i.e. that you
>>> are
>>> only considering part of the system which is defined by the reduced
>>> density matrix. The complete system of buckyball plus photons will
>>> show
>>> interference, even if the wavelength is small enough to resolve the
>>> slits provided you perform the right sort of measurement on the
>>> balls
>>> and photons.
>> That is false.
> 
> This is easy to see. Denote the buckyball state of a buckball moving through 
> the left slit by |L> and moving through the right slit by |R>. Suppose that a 
> photon is emitted by the by the buckyballs such that the ball moving through 
> the left slit emits a photon in a state |PL> that will be orthogonal to the 
> state |PR> of the photon emitted by the ball moving through the right slit . 
> The state of the system after the ball passes the slits is then:
> 
> |psi> = 1/sqrt(2) [|L>|PL> + |R>|PR>]
> 
> This state then evolves under unitary time evolution, we can write the state 
> just before the ball hits the screen as:
> 
> |psi_s> = 1/sqrt(2) [|L_s>|PL_s> + |R_s>|PR_s>]
> 
> There is then no interference patter on the screen for the buckyballs because 
> |PL_s> and |PR_s> are orthogonal, the unitary time evolution preserves the 
> orthogonality of the initial states. The probability to observe a buckyball 
> on position x on the screen is:
> 
> P(x) = ^2 = 1/2 [||^2 + ||^2] + Re[ 
> * ]
> 
> And the last interference term is zero because  = 0
> 
> But if we also observe the photon on another screen and keep the joint count 
> for buckyballs landing on spot x on the buckyball screen and for photons 
> landing on spot y on the photon screen as a function of x and y, then we do 
> have an interference pattern as a function of x for fixed y. If we de note by 
> U the unitary time evolution for the photons until they hit their screen, and 
> put |PL_t> =U|PL_s> and |PR_t> = U|PR_s>, then the probability distribution 
> is:
> 
> P(x,y) = ||^2 = 1/2 [||^2||^2 + 
> ||^2||^2] +Re[ * *]
> 
> The interference term Re[ * *] does not vanish 
> as it involves evaluating the components of  the buckyball and photon states 
> in the position basis and so there is no inner product involved anymore. For 
> fixed y the quantity * will have some value that will be 
> nonzero in general, so if we keep y fixed then there will be an interference 
> term.
> 
> So, we can conclude that invoking escaping IR photons does not male any sense 
> in this discussion because all it does is it scrambles the interference 
> pattern to make it invisible in a way that allows it to be recovered in 
> principle using measurements on those IR photons. You can, of course, erase 
> the interference patter by measuring the observable for the photons that has 
> |PR> and |PL> as its eigenstates. But even in that case the information will 
> still be there in the state of all the atoms of the measurement apparatus for 
> the photons. But if you don't perform any measurement then the information 
> will simply continue to exists in the escaping photons.
> 
> So, in general we can conclude by generalizing this to any large number of 
> particles that even with what we consider to be permanent records, you don't 
> get rid of the theoretical possibility of interference between the sectors 
> where those records are different. So, the existence of parallel worlds 
> cannot be made fully 100% irrelevant if QM is rigorously correct, and we 
> cannot therefore argue that QM is exactly equivalent to an alternative theory 
> that leaves out parallel worlds. Even though the difference may be almost 
> 100% insignificant FAPP, it's not exactly 100% even in the macroscopic realm.
> 
> The argument against the existence of

Re: Why are laws of physics stable?

> On 5 Jul 2021, at 22:19, Tomas Pales  wrote:
> 
> 
> On Monday, July 5, 2021 at 8:03:46 PM UTC+2 johnk...@gmail.com wrote:
> 
> How can my consciousness be located in a place that I am not conscious of?
> 
> You are conscious of certain parts of your brain (presumably those that have 
> high organized complexity of a certain kind), many of which are 
> representations of external objects.

External object are how universal number organise their continuation in 
arithmetic. There are no “external physical object” in the ontology.

With Mechanism, to refer to a physical world is no better than to refer to a 
god. It cannot work without adding non Turing emulable ability to the brain. It 
requites the abandon of Mechanism, and thus of Darwin.

With mechanism, on the contrary, Darwin type of explanation is extended up to 
the origin of the physical laws, through the first person consciousness 
selection of histories in arithmetic. That explains the many-world aspect of 
the physical reality, and the math confirms the quantum logical structures of 
the personally accessible histories, including why a subset is sharable among 
stable collection of machine.

The quantum is “just” the digital seen from inside. It is the many-worlds 
interpretation of arithmetic, made by almost all universal numbers (almost all 
= all except a finite number of exception, which are topologically isolated and 
of measure nul).

Bruno

> If there are no representations of your skull it is probably because they 
> have not been useful in evolution. It is more useful to be conscious of what 
> is going on around your body than of the interior of your skull.
> 
> We can't rule out that your consciousness is not located in your brain and 
> your brain just serves as a kind of preliminary processor of sensory data 
> that somehow sends the preprocessed data to some other object (soul?) where 
> your consciousness is located but there's not much evidence for that.
> 
> 
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Re: Why are laws of physics stable?


> On 5 Jul 2021, at 21:01, 'Brent Meeker' via Everything List 
>  wrote:
> 
> 
> On 7/5/2021 7:41 AM, John Clark wrote:
>> On Mon, Jul 5, 2021 at 9:44 AM Tomas Pales > > wrote:
>> 
>>  >> "Brain" is a noun, "consciousness" is not, that's why you can't measure 
>> consciousness by the pound or by the cubic inch.
>> 
>> > In English language it is used as a noun. Check out a dictionary:
>> 
>> consciousness noun 
>> I know, that's what my fourth grade teacher told me too, but I long-ago 
>> realized that neither she nor the lexicographers who wrote that big thick 
>> book are the fonts of all wisdom.
>> >> Intelligence is what a brain does not what a brain is, and because 
>> >> Darwinian Evolution is almost certainly correct, consciousness must be an 
>> >> inevitable byproduct of intelligence, therefore "consciousness" is not a 
>> >> noun, it's a word that describes what a noun (in this case the brain) 
>> >> does, in other words consciousness is an adjective.
>> 
>> > You mean a verb then, no?
>> 
>> I think adjective fits the bill a little better, I think Tomas Pales is the 
>> way atoms behave when they are arranged in a Tomaspalesian way.
>> 
>>  > consciousness is a spatiotemporal object.
>> 
>> I disagree, I think asking where my consciousness is located would be like 
>> asking where the number 11 or the color yellow  or "fast" is located.  If my 
>> brain is in Paris and I'm looking at a TV football game from Detroit and I'm 
>> listening to a friend in Australia on my telephone and I'm thinking about 
>> The Great Wall of China would it  make sense to say my consciousness is 
>> really located inside a box made of bone mounted on my shoulders when I have 
>> no conscious experience of being in a bone box on my shoulders? I don't 
>> think so.
> Yet a sharp blow to that bone box would eliminate your conscious experience 
> at least temporarily. 
> 
Only from the point of view of some conscious subject. From the point of view 
of the person associated to the brain in the box, that does not make sense, as 
it is associated to infinitely many truing universal relation.

The body is only a map on infinitely many histories. That can be proved both 
with QM-without-collapse, or in any non trivial combinatory algebra (like a 
model of arithmetic).


> So there's something there that is essential to your consciousness.
> 

What is “essential” are the infinitely many computations.

Since the 1930s we know that all computations are realised in any model (in the 
logician sense) of arithmetic, or of combinatory logic (Kxy = x, Sxyz = xz(yz)).

I know that this contradict 1492 years of materialist brainwashing, but 
“appearance of matter” are explained in arithmetic, and get contradictory when 
associated or singularised through any supplementary axioms, even the induction 
axioms used to define what an observer can be.

You assume some ontological commitment inconsistent with Mechanism here.


Bruno


> Brent
> 
>> 
>> John K ClarkSee what's on my new list at  Extropolis 
>> 
>> 
>> vxq0
>> 
>>  
>> 
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Re: Why are laws of physics stable?

> On 5 Jul 2021, at 15:03, John Clark  wrote:
> 
> On Mon, Jul 5, 2021 at 5:33 AM Tomas Pales  > wrote:
> 
> > I think consciousness is the brain,
> 
> I disagree.

Me too.

> "Brain" is a noun, "consciousness" is not,

I disagree with this too.

> that's why you can't measure consciousness by the pound or by the cubic inch. 
> Intelligence is what a brain does not what a brain is,

It could be a disposition. 

“What. Brain does” is ambiguous. Is it the neural firing, or the action of 
muscles, or … bombs and rockets.

> and because Darwinian Evolution is almost certainly correct,

Darwinism makes no explanative sense without Mechanism, but with Mechanism 
things are necessary like that:

NUMBER => COMPUTATION => CONSCIOUSNESS => PHYSICAL APPEARANCE => HUMAN 
CONSCIOUSNESS

> consciousness must be an inevitable byproduct of intelligence,

It is a relative number property (or combinator, program, digital machine, …). 
Each universal number in arithmetic is an initial consciousness starting point. 
The physical reality is the product of consciousness differentiation in the 
arithmetical reality (actually: in any model of any arithmetical or Turing 
universal theory).

> therefore "consciousness" is not a noun, it's a word that describes what a 
> noun (in this case the brain) does, in other words consciousness is an 
> adjective.

This is logically inconsistent with Descartes Mechanism. Without mechanism, it 
is consistent, but still problematic with Occam razor and the absence of 
evidence for primitive matter.

Bruno

> 
> John K ClarkSee what's on my new list at  Extropolis 
> 
> 
> ks77
> 
> 
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Re: Why are laws of physics stable?

2021-07-09 Thread Bruce Kellett

On Sat, Jul 10, 2021 at 10:21 AM smitra  wrote:

> On 08-07-2021 01:51, Bruce Kellett wrote:
> >
> > Do you dispute that that is what the paper by Hornberger et al. says?
> >
> > Bruce
>
> I don't dispute these results. The buckyballs are coming from a thermal
> reservoir at some finite temperature. We can avoid working with mixed
> states by simply considering the interference pattern for each pure
> state separately and then summing over the probability distribution over
> the pure states. But for this discussion we want to focus on what the
> interference pattern will be if all the buckyballs are in the same
> exited state. If we put the entire system ina finite volume then we have
> a countable set of allowed k-values for the photon momenta. We then have
> a set of allowed states for the photons defined by the allowed momenta
> and a polarization. We can then label these photon states using a number
> and then specify an arbitrary state for the photons by specifying how
> many photons we have in each state, and these numbers can then be equal
> to zero. If they are all zero then no photons are present.
>
> If only the right slit is open, the state of the buckyball and the
> photons just before the screen is hit can be denoted as:
>
> |Right> = sum over n1, n2,n3,...|R(n1,n2,n3,...)>|n1,n2,n3,..>
>
> where |R(n1,n2,...)> denotes the quantum state of the buckyball if it
> emits n1 photons in state 1, n2 photons in state 2 etc. The state of the
> photons is then denoted as |n1,n2,n3,..>
>
>
> If only the left slit is open, the state of the buckyball and the
> photons just before the screen is hit can be denoted as:
>
> |Left> = sum over n1, n2,n3,...|L(n1,n2,n3,...)>|n1,n2,n3,..>
>
> where |R(n1,n2,...)> denotes the quantum state of the buckyball. Here we
> note that the state of the photons will pick up a phase factor relative
> to the case of only the right slit being open, but we can then absorb
> this phase factor in |L(n1,n2,n3,...)>.
>
> With both slits open, we'll then have a state of the form:
>
> |psi> = 1/sqrt(2) [|Right> + |Left>]
>
> The inner product of |psi> with some position eigenstate |x>,  is
> then a state vector for the photon states, the squared norm of that
> state vector is the probability of finding the buckyball at position x,
> because this is the sum over the probabilities for photons over all
> possible photon states. So, the probability is then:
>
> P(x) = ^2 = 1/2 [^2 + ^2] +
> Re[]
>
>
> We can then evaluate the interference term as follows:
>
> Re[] = Re sum over n1, n2,n3,...m1,m2,m3
> *
>
> Using that  = 0 unless m_j = n_j for all
> j, in which case this inner product equals 1, we then have:
>
> Re[] = Re sum over n1, n2,n3
> * =
>
>
> Re[*] +
> Re[*] + .  (1)
>
> As explained above when there are photons present then we've absorbed
> the phase factor due to translation of the photon states in the states
> |L(n1,n2,n3...)>. For each wave vector k there is a factor for each
> photon with that wavevector of exp(i k dot r) where r is the position of
> the left slit w.r.t. the right slit. So, the total phase factor will be:
>
> Product over j of exp(i kj nj dot r)
>
> In the experiment there is then an additional summation over the pure
> states of the buckyballs. If the temperature is low then the summation
> will consist of states for which |R(0,0,0,..> and |L(0,0,0,..> are the
> dominant terms, as most of the time no photons will be emitted. At
> higher temperatures the typical states there will be contributions from
> different numbers pf photons, so the interference pattern will be a sum
> of many different terms in (1) with comparable norms, they come with
> different phase factors due to the different numbers of photons with
> different momenta. So, the interference pattern will be washed out.
>


This analysis contradicts what you said in your first analysis. In the
first analysis, you claimed that no interference would be seen if the IR
photons were not detected. You seem to have dropped this notion in the
above. You now say that there is interference when no photons are present,
but this is washed out when there are different numbers of photons.

I think you are making the same basic mistake that you made when we
previously discussed  two-slit interference: You are analysing the two slit
case as a sum over single slit patterns, and assuming the emission of
independent photons from the ball through each slit. The trouble is that
there are not two balls, one for each slit. The same ball goes through both
slits and there is only one ball emitting (or not emitting) photons.

Neither of your analyses actually explain the observed behaviour.

Bruce

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Re: Why are laws of physics stable?

2021-07-09 Thread 'Brent Meeker' via Everything List




On 7/9/2021 4:40 PM, Lawrence Crowell wrote:



On Wednesday, July 7, 2021 at 9:29:24 PM UTC-5 Brent wrote:


On 7/7/2021 2:04 PM, Jason Resch wrote:



On Wed, Jul 7, 2021 at 3:43 PM 'Brent Meeker' via Everything List
 wrote:


On 7/7/2021 10:09 AM, Jason Resch wrote:



On Wed, Jul 7, 2021 at 11:53 AM 'Brent Meeker' via
Everything List  wrote:


On 7/7/2021 2:24 AM, Jason Resch wrote:



On Wed, Jul 7, 2021, 12:14 AM 'Brent Meeker' via
Everything List  wrote:


On 7/6/2021 6:50 PM, Jason Resch wrote:



On Tue, Jul 6, 2021, 9:39 PM Bruce Kellett
 wrote:

On Wed, Jul 7, 2021 at 11:29 AM Jason Resch
 wrote:

On Tue, Jul 6, 2021, 4:07 PM 'Brent
Meeker' via Everything List
 wrote:

On 7/6/2021 10:34 AM, Jason Resch wrote:


On Tue, Jul 6, 2021 at 12:27 PM
'Brent Meeker' via Everything List
 wrote:

And you're never going to find a
being that behaves intelligently
based on information that can be
quantum erased.

You need only a quantum computer with
enough qubits.


Can you prove that? How does this
quantum intelligence ever arrive at a
definite decision?


Prove? No. But I think I can justify it:

1. Quantum computers are Turing
equivalent, they can compute anything a
classical computer can.

2. Human brains are believed to operate
according to physical laws, all known of
which are computable.

3. Humans are conscious.

4. By any of: Chalmers's principle of
"Organizational invariance", or "multiple
realizability", or the "Generalized
Anti-Zombie Principle", or the
"computational theory of mind", a
functionally equivalent computation to
that of a conscious human brain will be
equivalently conscious to that brain.

5. Quantum computers are reversible.

By 1 & 2, a quantum computer can simulate
a human brain. By 3 & 4, such an emulation
will be conscious. By 5 any computation
performed by a quantum computer can be
quantum erased by reversing the circuit
back to its starting state.

It reaches a definite decision by virtue
of completing its processing before
ultimately being reversed. This prevents
an outside observer from learning the
decision, but it's made nonetheless during
the course of the processing.


How do you know that it has reached a definite
decision? Without having it print out some
irreversible record? If it prints out a
(pseudo-)classical record, the initial state
is not recoverable.

Bruce


By either:

1. Analyzing the circuit


But the question is whether such a circuit is possible.

Do you disagree with any of the five premises I defined
above? If not do you see a flaw in my reasoning or
conclusions? If not, then why shouldn't such a circuit
be possible?


This what I find dubious: /"It reaches a definite
decision by virtue of completing its processing before
ultimately being reversed. This prevents an outside
observer from learning the decision, but it's made
nonetheless during the course of the processing." /
First, I doubt that it both reach a definite decision
and have that quantum erasable.

If you doubt it reaches a certain definite decision state,
you could interrupt the quantum computer midway through its
processing and entangle yourself with one of its superposed
states to verify that the AI/mind was in a state of having
reached a definition conclusion.



?? If I do that by entangling with a superposition, then I
either collapse it or "I'm of

Re: Why are laws of physics stable?

2021-07-09 Thread smitra


On 08-07-2021 01:51, Bruce Kellett wrote:

On Wed, Jul 7, 2021 at 5:26 PM smitra  wrote:


On 07-07-2021 01:17, Bruce Kellett wrote:




Your idea of QM is sadly flawed. The real professional quantum
analysis given in the quoted paper shows how the observed effects

are

completely consistent with quantum mechanics. The emission of

thermal

radiation by the heated balls leads to a clear and evident loss of
coherence. Your pseudo-analysis has nothing to do with either

quantum

mechanics or the actual set-up of this buckyball experiment.


As I said, what I wrote is 100% consistent with their results.


Your analysis, as I understand it, suggests that if the IR photons are
not observed, no interference pattern is seen. This seems to overlook
the fact that the photons are emitted only probabilistically -- there
need not be any photons at all, and yet the experiment sees a
buckyball interference pattern at the lower temperatures. This pattern
is gradually washed out as the temperature of the balls is increased.
No photons are ever detected in that experiment.

Do you dispute that that is what the paper by Hornberger et al. says?

Bruce


I don't dispute these results. The buckyballs are coming from a thermal 
reservoir at some finite temperature. We can avoid working with mixed 
states by simply considering the interference pattern for each pure 
state separately and then summing over the probability distribution over 
the pure states. But for this discussion we want to focus on what the 
interference pattern will be if all the buckyballs are in the same 
exited state. If we put the entire system ina finite volume then we have 
a countable set of allowed k-values for the photon momenta. We then have 
a set of allowed states for the photons defined by the allowed momenta 
and a polarization. We can then label these photon states using a number 
and then specify an arbitrary state for the photons by specifying how 
many photons we have in each state, and these numbers can then be equal 
to zero. If they are all zero then no photons are present.


If only the right slit is open, the state of the buckyball and the 
photons just before the screen is hit can be denoted as:


|Right> = sum over n1, n2,n3,...|R(n1,n2,n3,...)>|n1,n2,n3,..>

where |R(n1,n2,...)> denotes the quantum state of the buckyball if it 
emits n1 photons in state 1, n2 photons in state 2 etc. The state of the 
photons is then denoted as |n1,n2,n3,..>



If only the left slit is open, the state of the buckyball and the 
photons just before the screen is hit can be denoted as:


|Left> = sum over n1, n2,n3,...|L(n1,n2,n3,...)>|n1,n2,n3,..>

where |R(n1,n2,...)> denotes the quantum state of the buckyball. Here we 
note that the state of the photons will pick up a phase factor relative 
to the case of only the right slit being open, but we can then absorb 
this phase factor in |L(n1,n2,n3,...)>.


With both slits open, we'll then have a state of the form:

|psi> = 1/sqrt(2) [|Right> + |Left>]

The inner product of |psi> with some position eigenstate |x>,  is 
then a state vector for the photon states, the squared norm of that 
state vector is the probability of finding the buckyball at position x, 
because this is the sum over the probabilities for photons over all 
possible photon states. So, the probability is then:


P(x) = ^2 = 1/2 [^2 + ^2] + 
Re[]



We can then evaluate the interference term as follows:

Re[] = Re sum over n1, n2,n3,...m1,m2,m3 
*


Using that  = 0 unless m_j = n_j for all 
j, in which case this inner product equals 1, we then have:


Re[] = Re sum over n1, n2,n3 
* =



Re[*] + 
Re[*] + .  (1)


As explained above when there are photons present then we've absorbed 
the phase factor due to translation of the photon states in the states 
|L(n1,n2,n3...)>. For each wave vector k there is a factor for each 
photon with that wavevector of exp(i k dot r) where r is the position of 
the left slit w.r.t. the right slit. So, the total phase factor will be:


Product over j of exp(i kj nj dot r)

In the experiment there is then an additional summation over the pure 
states of the buckyballs. If the temperature is low then the summation 
will consist of states for which |R(0,0,0,..> and |L(0,0,0,..> are the 
dominant terms, as most of the time no photons will be emitted. At 
higher temperatures the typical states there will be contributions from 
different numbers pf photons, so the interference pattern will be a sum 
of many different terms in (1) with comparable norms, they come with 
different phase factors due to the different numbers of photons with 
different momenta. So, the interference pattern will be washed out.


Saibal

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Re: Why are laws of physics stable?

2021-07-09 Thread Lawrence Crowell



On Wednesday, July 7, 2021 at 9:29:24 PM UTC-5 Brent wrote:

>
> On 7/7/2021 2:04 PM, Jason Resch wrote:
>
>
>
> On Wed, Jul 7, 2021 at 3:43 PM 'Brent Meeker' via Everything List <
> everyth...@googlegroups.com> wrote:
>
>>
>> On 7/7/2021 10:09 AM, Jason Resch wrote:
>>
>>
>>
>> On Wed, Jul 7, 2021 at 11:53 AM 'Brent Meeker' via Everything List <
>> everyth...@googlegroups.com> wrote:
>>
>>>
>>> On 7/7/2021 2:24 AM, Jason Resch wrote:
>>>
>>>
>>>
>>> On Wed, Jul 7, 2021, 12:14 AM 'Brent Meeker' via Everything List <
>>> everyth...@googlegroups.com> wrote:
>>>

 On 7/6/2021 6:50 PM, Jason Resch wrote:



 On Tue, Jul 6, 2021, 9:39 PM Bruce Kellett  wrote:

> On Wed, Jul 7, 2021 at 11:29 AM Jason Resch  
> wrote:
>
>> On Tue, Jul 6, 2021, 4:07 PM 'Brent Meeker' via Everything List <
>> everyth...@googlegroups.com> wrote:
>>
>>> On 7/6/2021 10:34 AM, Jason Resch wrote:
>>>
>>> On Tue, Jul 6, 2021 at 12:27 PM 'Brent Meeker' via Everything List <
>>> everyth...@googlegroups.com> wrote:
>>>
 And you're never going to find a being that behaves intelligently 
 based on information that can be quantum erased.

>>> You need only a quantum computer with enough qubits.
>>>
>>> Can you prove that?  How does this quantum intelligence ever arrive 
>>> at a definite decision?
>>>
>>
>> Prove? No. But I think I can justify it:
>>
>> 1. Quantum computers are Turing equivalent, they can compute anything 
>> a classical computer can.
>>
>> 2. Human brains are believed to operate according to physical laws, 
>> all known of which are computable.
>>
>> 3. Humans are conscious.
>>
>> 4. By any of: Chalmers's principle of "Organizational invariance", or 
>> "multiple realizability", or the "Generalized Anti-Zombie Principle", or 
>> the "computational theory of mind", a functionally equivalent 
>> computation 
>> to that of a conscious human brain will be equivalently conscious to 
>> that 
>> brain.
>>
>> 5. Quantum computers are reversible.
>>
>> By 1 & 2, a quantum computer can simulate a human brain. By 3 & 4, 
>> such an emulation will be conscious. By 5 any computation performed by a 
>> quantum computer can be quantum erased by reversing the circuit back to 
>> its 
>> starting state.
>>
>> It reaches a definite decision by virtue of completing its processing 
>> before ultimately being reversed. This prevents an outside observer from 
>> learning the decision, but it's made nonetheless during the course of 
>> the 
>> processing.
>>
>
> How do you know that it has reached a definite decision? Without 
> having it print out some irreversible record? If it prints out a 
> (pseudo-)classical record, the initial state is not recoverable.
>
> Bruce
>

 By either:

 1. Analyzing the circuit

 But the question is whether such a circuit is possible.

>>> Do you disagree with any of the five premises I defined above? If not do 
>>> you see a flaw in my reasoning or conclusions? If not, then why shouldn't 
>>> such a circuit be possible?
>>>
>>> This what I find dubious: *"It reaches a definite decision by virtue of 
>>> completing its processing before ultimately being reversed. This prevents 
>>> an outside observer from learning the decision, but it's made nonetheless 
>>> during the course of the processing." * First, I doubt that it both 
>>> reach a definite decision and have that quantum erasable. 
>>>
>> If you doubt it reaches a certain definite decision state, you could 
>> interrupt the quantum computer midway through its processing and entangle 
>> yourself with one of its superposed states to verify that the AI/mind was 
>> in a state of having reached a definition conclusion.
>>
>>
>> ?? If I do that by entangling with a superposition, then I either 
>> collapse it or "I'm of two minds".
>>
>
> Yeah you spoil the process by interrupting it early, but it lets you 
> verify the computation reaches those intermediate states in the course of 
> its normal evolution, including in those that you allow the algorithm to 
> run to completion.
>  
>
>>
>>  
>>
>>> Second, you've made "decision" something internal.  Intelligence 
>>> requires acting in the world.
>>>
>>>
>>> The environment for this AI are the qubits initialized as the input to 
>> the mind. It acts in this world by performing actions that ultimately 
>> affect the output of this quantum computation.
>>
>>
>> My original point was, "And you're never going to find a being that 
>> behaves intelligently based on information that can be quantum erased."  In 
>> the environment A=0, B=0, and any other set of A, B values the algorithm 
>> outputs B=1 and then erases it.  Is this intelligent behavior?
>>
> It's perhaps a thermostat level of intelligence, but you can

Re: Why are laws of physics stable?

2021-07-07 Thread 'Brent Meeker' via Everything List



On 7/7/2021 2:04 PM, Jason Resch wrote:



On Wed, Jul 7, 2021 at 3:43 PM 'Brent Meeker' via Everything List 
> wrote:



On 7/7/2021 10:09 AM, Jason Resch wrote:



On Wed, Jul 7, 2021 at 11:53 AM 'Brent Meeker' via Everything
List mailto:everything-list@googlegroups.com>> wrote:


On 7/7/2021 2:24 AM, Jason Resch wrote:



On Wed, Jul 7, 2021, 12:14 AM 'Brent Meeker' via Everything
List mailto:everything-list@googlegroups.com>> wrote:


On 7/6/2021 6:50 PM, Jason Resch wrote:



On Tue, Jul 6, 2021, 9:39 PM Bruce Kellett
mailto:bhkellet...@gmail.com>>
wrote:

On Wed, Jul 7, 2021 at 11:29 AM Jason Resch
mailto:jasonre...@gmail.com>> wrote:

On Tue, Jul 6, 2021, 4:07 PM 'Brent Meeker' via
Everything List
mailto:everything-list@googlegroups.com>> wrote:

On 7/6/2021 10:34 AM, Jason Resch wrote:


On Tue, Jul 6, 2021 at 12:27 PM 'Brent
Meeker' via Everything List
mailto:everything-list@googlegroups.com>>
wrote:

And you're never going to find a being
that behaves intelligently based on
information that can be quantum erased.

You need only a quantum computer with
enough qubits.


Can you prove that? How does this quantum
intelligence ever arrive at a definite
decision?


Prove? No. But I think I can justify it:

1. Quantum computers are Turing equivalent,
they can compute anything a classical computer can.

2. Human brains are believed to operate
according to physical laws, all known of which
are computable.

3. Humans are conscious.

4. By any of: Chalmers's principle of
"Organizational invariance", or "multiple
realizability", or the "Generalized Anti-Zombie
Principle", or the "computational theory of
mind", a functionally equivalent computation to
that of a conscious human brain will be
equivalently conscious to that brain.

5. Quantum computers are reversible.

By 1 & 2, a quantum computer can simulate a
human brain. By 3 & 4, such an emulation will
be conscious. By 5 any computation performed by
a quantum computer can be quantum erased by
reversing the circuit back to its starting state.

It reaches a definite decision by virtue of
completing its processing before ultimately
being reversed. This prevents an outside
observer from learning the decision, but it's
made nonetheless during the course of the
processing.


How do you know that it has reached a definite
decision? Without having it print out some
irreversible record? If it prints out a
(pseudo-)classical record, the initial state is not
recoverable.

Bruce


By either:

1. Analyzing the circuit


But the question is whether such a circuit is possible.

Do you disagree with any of the five premises I defined
above? If not do you see a flaw in my reasoning or
conclusions? If not, then why shouldn't such a circuit be
possible?


This what I find dubious: /"It reaches a definite decision by
virtue of completing its processing before ultimately being
reversed. This prevents an outside observer from learning the
decision, but it's made nonetheless during the course of the
processing." / First, I doubt that it both reach a definite
decision and have that quantum erasable.

If you doubt it reaches a certain definite decision state, you
could interrupt the quantum computer midway through its
processing and entangle yourself with one of its superposed
states to verify that the AI/mind was in a state of having
reached a definition conclusion.



?? If I do that by entangling with a superposition, then I either
collapse it or "I'm of two minds".


Yeah you spoil the process by interrupting it early, but it lets you 
verify the computation reaches those intermediate states in the course 
of its normal evolution, including in those that you allow the 
algorithm to run to

Re: Why are laws of physics stable?

2021-07-07 Thread Bruce Kellett

On Wed, Jul 7, 2021 at 5:26 PM smitra  wrote:

> On 07-07-2021 01:17, Bruce Kellett wrote:
>
>
> > Your idea of QM is sadly flawed. The real professional quantum
> > analysis given in the quoted paper shows how the observed effects are
> > completely consistent with quantum mechanics. The emission of thermal
> > radiation by the heated balls leads to a clear and evident loss of
> > coherence. Your pseudo-analysis has nothing to do with either quantum
> > mechanics or the actual set-up of this buckyball experiment.
>
> As I said, what I wrote is 100% consistent with their results.

Your analysis, as I understand it, suggests that if the IR photons are not
observed, no interference pattern is seen. This seems to overlook the fact
that the photons are emitted only probabilistically -- there need not be
any photons at all, and yet the experiment sees a buckyball
interference pattern at the lower temperatures. This pattern is gradually
washed out as the temperature of the balls is increased. No photons are
ever detected in that experiment.

Do you dispute that that is what the paper by Hornberger et al. says?

Bruce

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Re: Why are laws of physics stable?

2021-07-07 Thread Jason Resch

On Wed, Jul 7, 2021 at 3:43 PM 'Brent Meeker' via Everything List <
everything-list@googlegroups.com> wrote:

>
> On 7/7/2021 10:09 AM, Jason Resch wrote:
>
>
>
> On Wed, Jul 7, 2021 at 11:53 AM 'Brent Meeker' via Everything List <
> everything-list@googlegroups.com> wrote:
>
>>
>> On 7/7/2021 2:24 AM, Jason Resch wrote:
>>
>>
>>
>> On Wed, Jul 7, 2021, 12:14 AM 'Brent Meeker' via Everything List <
>> everything-list@googlegroups.com> wrote:
>>
>>>
>>> On 7/6/2021 6:50 PM, Jason Resch wrote:
>>>
>>>
>>>
>>> On Tue, Jul 6, 2021, 9:39 PM Bruce Kellett 
>>> wrote:
>>>
 On Wed, Jul 7, 2021 at 11:29 AM Jason Resch 
 wrote:

> On Tue, Jul 6, 2021, 4:07 PM 'Brent Meeker' via Everything List <
> everything-list@googlegroups.com> wrote:
>
>> On 7/6/2021 10:34 AM, Jason Resch wrote:
>>
>> On Tue, Jul 6, 2021 at 12:27 PM 'Brent Meeker' via Everything List <
>> everything-list@googlegroups.com> wrote:
>>
>>> And you're never going to find a being that behaves intelligently
>>> based on information that can be quantum erased.
>>>
>> You need only a quantum computer with enough qubits.
>>
>> Can you prove that?  How does this quantum intelligence ever arrive
>> at a definite decision?
>>
>
> Prove? No. But I think I can justify it:
>
> 1. Quantum computers are Turing equivalent, they can compute anything
> a classical computer can.
>
> 2. Human brains are believed to operate according to physical laws,
> all known of which are computable.
>
> 3. Humans are conscious.
>
> 4. By any of: Chalmers's principle of "Organizational invariance", or
> "multiple realizability", or the "Generalized Anti-Zombie Principle", or
> the "computational theory of mind", a functionally equivalent computation
> to that of a conscious human brain will be equivalently conscious to that
> brain.
>
> 5. Quantum computers are reversible.
>
> By 1 & 2, a quantum computer can simulate a human brain. By 3 & 4,
> such an emulation will be conscious. By 5 any computation performed by a
> quantum computer can be quantum erased by reversing the circuit back to 
> its
> starting state.
>
> It reaches a definite decision by virtue of completing its processing
> before ultimately being reversed. This prevents an outside observer from
> learning the decision, but it's made nonetheless during the course of the
> processing.
>

 How do you know that it has reached a definite decision? Without having
 it print out some irreversible record? If it prints out a
 (pseudo-)classical record, the initial state is not recoverable.

 Bruce

>>>
>>> By either:
>>>
>>> 1. Analyzing the circuit
>>>
>>> But the question is whether such a circuit is possible.
>>>
>> Do you disagree with any of the five premises I defined above? If not do
>> you see a flaw in my reasoning or conclusions? If not, then why shouldn't
>> such a circuit be possible?
>>
>> This what I find dubious: *"It reaches a definite decision by virtue of
>> completing its processing before ultimately being reversed. This prevents
>> an outside observer from learning the decision, but it's made nonetheless
>> during the course of the processing." * First, I doubt that it both
>> reach a definite decision and have that quantum erasable.
>>
> If you doubt it reaches a certain definite decision state, you could
> interrupt the quantum computer midway through its processing and entangle
> yourself with one of its superposed states to verify that the AI/mind was
> in a state of having reached a definition conclusion.
>
>
> ?? If I do that by entangling with a superposition, then I either collapse
> it or "I'm of two minds".
>

Yeah you spoil the process by interrupting it early, but it lets you verify
the computation reaches those intermediate states in the course of its
normal evolution, including in those that you allow the algorithm to run to
completion.


>
>
>
>> Second, you've made "decision" something internal.  Intelligence requires
>> acting in the world.
>>
>>
>> The environment for this AI are the qubits initialized as the input to
> the mind. It acts in this world by performing actions that ultimately
> affect the output of this quantum computation.
>
>
> My original point was, "And you're never going to find a being that
> behaves intelligently based on information that can be quantum erased."  In
> the environment A=0, B=0, and any other set of A, B values the algorithm
> outputs B=1 and then erases it.  Is this intelligent behavior?
>
It's perhaps a thermostat level of intelligence, but you can make it
arbitrarily complex, as in Deutsch's AI example that does the same thing as
this simple circuit.

Jason



> Brent
>
>
>
>
>>
>>
>> 2. Having the circuit do something useful and verifiable (as in my
>>> factoring example)
>>>
>>> How would you know that had a causal connection to

Re: Why are laws of physics stable?

2021-07-07 Thread 'Brent Meeker' via Everything List



On 7/7/2021 10:09 AM, Jason Resch wrote:



On Wed, Jul 7, 2021 at 11:53 AM 'Brent Meeker' via Everything List 
> wrote:



On 7/7/2021 2:24 AM, Jason Resch wrote:



On Wed, Jul 7, 2021, 12:14 AM 'Brent Meeker' via Everything List
mailto:everything-list@googlegroups.com>> wrote:


On 7/6/2021 6:50 PM, Jason Resch wrote:



On Tue, Jul 6, 2021, 9:39 PM Bruce Kellett
mailto:bhkellet...@gmail.com>> wrote:

On Wed, Jul 7, 2021 at 11:29 AM Jason Resch
mailto:jasonre...@gmail.com>> wrote:

On Tue, Jul 6, 2021, 4:07 PM 'Brent Meeker' via
Everything List mailto:everything-list@googlegroups.com>> wrote:

On 7/6/2021 10:34 AM, Jason Resch wrote:


On Tue, Jul 6, 2021 at 12:27 PM 'Brent Meeker'
via Everything List
mailto:everything-list@googlegroups.com>> wrote:

And you're never going to find a being that
behaves intelligently based on information
that can be quantum erased.

You need only a quantum computer with enough
qubits.


Can you prove that?  How does this quantum
intelligence ever arrive at a definite decision?


Prove? No. But I think I can justify it:

1. Quantum computers are Turing equivalent, they can
compute anything a classical computer can.

2. Human brains are believed to operate according to
physical laws, all known of which are computable.

3. Humans are conscious.

4. By any of: Chalmers's principle of
"Organizational invariance", or "multiple
realizability", or the "Generalized Anti-Zombie
Principle", or the "computational theory of mind", a
functionally equivalent computation to that of a
conscious human brain will be equivalently conscious
to that brain.

5. Quantum computers are reversible.

By 1 & 2, a quantum computer can simulate a human
brain. By 3 & 4, such an emulation will be
conscious. By 5 any computation performed by a
quantum computer can be quantum erased by reversing
the circuit back to its starting state.

It reaches a definite decision by virtue of
completing its processing before ultimately being
reversed. This prevents an outside observer from
learning the decision, but it's made nonetheless
during the course of the processing.


How do you know that it has reached a definite decision?
Without having it print out some irreversible record? If
it prints out a (pseudo-)classical record, the initial
state is not recoverable.

Bruce


By either:

1. Analyzing the circuit


But the question is whether such a circuit is possible.

Do you disagree with any of the five premises I defined above? If
not do you see a flaw in my reasoning or conclusions? If not,
then why shouldn't such a circuit be possible?


This what I find dubious: /"It reaches a definite decision by
virtue of completing its processing before ultimately being
reversed. This prevents an outside observer from learning the
decision, but it's made nonetheless during the course of the
processing." / First, I doubt that it both reach a definite
decision and have that quantum erasable.

If you doubt it reaches a certain definite decision state, you could 
interrupt the quantum computer midway through its processing and 
entangle yourself with one of its superposed states to verify that the 
AI/mind was in a state of having reached a definition conclusion.



?? If I do that by entangling with a superposition, then I either 
collapse it or "I'm of two minds".




Second, you've made "decision" something internal. Intelligence
requires acting in the world.


The environment for this AI are the qubits initialized as the input to 
the mind. It acts in this world by performing actions that ultimately 
affect the output of this quantum computation.



My original point was, "And you're never going to find a being that 
behaves intelligently based on information that can be quantum erased."  
In the environment A=0, B=0, and any other set of A, B values the 
algorithm outputs B=1 and then erases it.  Is this intelligent behavior?


Brent






2. Having the circuit do something useful and verifiable (as
in my factoring example)


How would you know that had a causal connection to the
quantum erasable knowledge?

This is why I had

Re: Why are laws of physics stable?

2021-07-07 Thread Jason Resch

On Wed, Jul 7, 2021 at 11:53 AM 'Brent Meeker' via Everything List <
everything-list@googlegroups.com> wrote:

>
> On 7/7/2021 2:24 AM, Jason Resch wrote:
>
>
>
> On Wed, Jul 7, 2021, 12:14 AM 'Brent Meeker' via Everything List <
> everything-list@googlegroups.com> wrote:
>
>>
>> On 7/6/2021 6:50 PM, Jason Resch wrote:
>>
>>
>>
>> On Tue, Jul 6, 2021, 9:39 PM Bruce Kellett  wrote:
>>
>>> On Wed, Jul 7, 2021 at 11:29 AM Jason Resch 
>>> wrote:
>>>
 On Tue, Jul 6, 2021, 4:07 PM 'Brent Meeker' via Everything List <
 everything-list@googlegroups.com> wrote:

> On 7/6/2021 10:34 AM, Jason Resch wrote:
>
> On Tue, Jul 6, 2021 at 12:27 PM 'Brent Meeker' via Everything List <
> everything-list@googlegroups.com> wrote:
>
>> And you're never going to find a being that behaves intelligently
>> based on information that can be quantum erased.
>>
> You need only a quantum computer with enough qubits.
>
> Can you prove that?  How does this quantum intelligence ever arrive at
> a definite decision?
>

 Prove? No. But I think I can justify it:

 1. Quantum computers are Turing equivalent, they can compute anything a
 classical computer can.

 2. Human brains are believed to operate according to physical laws, all
 known of which are computable.

 3. Humans are conscious.

 4. By any of: Chalmers's principle of "Organizational invariance", or
 "multiple realizability", or the "Generalized Anti-Zombie Principle", or
 the "computational theory of mind", a functionally equivalent computation
 to that of a conscious human brain will be equivalently conscious to that
 brain.

 5. Quantum computers are reversible.

 By 1 & 2, a quantum computer can simulate a human brain. By 3 & 4, such
 an emulation will be conscious. By 5 any computation performed by a quantum
 computer can be quantum erased by reversing the circuit back to its
 starting state.

 It reaches a definite decision by virtue of completing its processing
 before ultimately being reversed. This prevents an outside observer from
 learning the decision, but it's made nonetheless during the course of the
 processing.

>>>
>>> How do you know that it has reached a definite decision? Without having
>>> it print out some irreversible record? If it prints out a
>>> (pseudo-)classical record, the initial state is not recoverable.
>>>
>>> Bruce
>>>
>>
>> By either:
>>
>> 1. Analyzing the circuit
>>
>> But the question is whether such a circuit is possible.
>>
> Do you disagree with any of the five premises I defined above? If not do
> you see a flaw in my reasoning or conclusions? If not, then why shouldn't
> such a circuit be possible?
>
> This what I find dubious: *"It reaches a definite decision by virtue of
> completing its processing before ultimately being reversed. This prevents
> an outside observer from learning the decision, but it's made nonetheless
> during the course of the processing." * First, I doubt that it both reach
> a definite decision and have that quantum erasable.
>
If you doubt it reaches a certain definite decision state, you could
interrupt the quantum computer midway through its processing and entangle
yourself with one of its superposed states to verify that the AI/mind was
in a state of having reached a definition conclusion.



> Second, you've made "decision" something internal.  Intelligence requires
> acting in the world.
>
>
> The environment for this AI are the qubits initialized as the input to the
mind. It acts in this world by performing actions that ultimately affect
the output of this quantum computation.


>
>
> 2. Having the circuit do something useful and verifiable (as in my
>> factoring example)
>>
>> How would you know that had a causal connection to the quantum erasable
>> knowledge?
>>
> This is why I had the information pass through the "AI function" before
> being used in Shor's algorithm. That way there was a causal connection with
> the result that would be communicated to the outside world.
>
>
> 3. Having the circuit output that it did observe a definite value but
>> without reporting which value it observed (as in Deutsch's original example)
>>
>> Again, how do you know such a circuit is possible?   Most quantum
>> computations only produce probable answers in a decohered readout.
>>
> Ignoring the AI aspect this is a simple and non probabilistic circuit:
>
> 1. Initialize qubit A to 0
> 2. Initialize qubit B to 0
> 3. Put qubit A into superposition of (0 and 1) via Hadamard gate
> 4. Apply Controlled NOT gate to (A, B) using A as the control bit to
> read/copy the bit value of A to the state of B. (B now has a definite value
> of 1 OR it has a definite value of 0)
> 5. Apply Pauli-X (NOT gate) to (B) to flip the bit value of B (it is now
> opposite of A).
> 6. Apply Controlled NOT gate to (A, B), to have the effect of computing (A
> XOR B)

Re: Why are laws of physics stable?

2021-07-07 Thread 'Brent Meeker' via Everything List



On 7/7/2021 2:24 AM, Jason Resch wrote:



On Wed, Jul 7, 2021, 12:14 AM 'Brent Meeker' via Everything List 
> wrote:



On 7/6/2021 6:50 PM, Jason Resch wrote:



On Tue, Jul 6, 2021, 9:39 PM Bruce Kellett mailto:bhkellet...@gmail.com>> wrote:

On Wed, Jul 7, 2021 at 11:29 AM Jason Resch
mailto:jasonre...@gmail.com>> wrote:

On Tue, Jul 6, 2021, 4:07 PM 'Brent Meeker' via
Everything List mailto:everything-list@googlegroups.com>> wrote:

On 7/6/2021 10:34 AM, Jason Resch wrote:


On Tue, Jul 6, 2021 at 12:27 PM 'Brent Meeker' via
Everything List mailto:everything-list@googlegroups.com>> wrote:

And you're never going to find a being that
behaves intelligently based on information that
can be quantum erased.

You need only a quantum computer with enough qubits.


Can you prove that?  How does this quantum
intelligence ever arrive at a definite decision?


Prove? No. But I think I can justify it:

1. Quantum computers are Turing equivalent, they can
compute anything a classical computer can.

2. Human brains are believed to operate according to
physical laws, all known of which are computable.

3. Humans are conscious.

4. By any of: Chalmers's principle of "Organizational
invariance", or "multiple realizability", or the
"Generalized Anti-Zombie Principle", or the
"computational theory of mind", a functionally equivalent
computation to that of a conscious human brain will be
equivalently conscious to that brain.

5. Quantum computers are reversible.

By 1 & 2, a quantum computer can simulate a human brain.
By 3 & 4, such an emulation will be conscious. By 5 any
computation performed by a quantum computer can be
quantum erased by reversing the circuit back to its
starting state.

It reaches a definite decision by virtue of completing
its processing before ultimately being reversed. This
prevents an outside observer from learning the decision,
but it's made nonetheless during the course of the
processing.


How do you know that it has reached a definite decision?
Without having it print out some irreversible record? If it
prints out a (pseudo-)classical record, the initial state is
not recoverable.

Bruce


By either:

1. Analyzing the circuit


But the question is whether such a circuit is possible.

Do you disagree with any of the five premises I defined above? If not 
do you see a flaw in my reasoning or conclusions? If not, then why 
shouldn't such a circuit be possible?


This what I find dubious: /"It reaches a definite decision by virtue of 
completing its processing before ultimately being reversed. This 
prevents an outside observer from learning the decision, but it's made 
nonetheless during the course of the processing." / First, I doubt that 
it both reach a definite decision and have that quantum erasable.  
Second, you've made "decision" something internal.  Intelligence 
requires acting in the world.







2. Having the circuit do something useful and verifiable (as in
my factoring example)


How would you know that had a causal connection to the quantum
erasable knowledge?

This is why I had the information pass through the "AI function" 
before being used in Shor's algorithm. That way there was a causal 
connection with the result that would be communicated to the outside 
world.




3. Having the circuit output that it did observe a definite value
but without reporting which value it observed (as in Deutsch's
original example)


Again, how do you know such a circuit is possible? Most quantum
computations only produce probable answers in a decohered readout.

Ignoring the AI aspect this is a simple and non probabilistic circuit:

1. Initialize qubit A to 0
2. Initialize qubit B to 0
3. Put qubit A into superposition of (0 and 1) via Hadamard gate
4. Apply Controlled NOT gate to (A, B) using A as the control bit to 
read/copy the bit value of A to the state of B. (B now has a definite 
value of 1 OR it has a definite value of 0)
5. Apply Pauli-X (NOT gate) to (B) to flip the bit value of B (it is 
now opposite of A).
6. Apply Controlled NOT gate to (A, B), to have the effect of 
computing (A XOR B) and storing the result in B. If A was measured in 
step 4 as 0, B will now be 1. Otherwise, if A was measured in step 4 
to be 1, B will now be 1. We now have evidence in qubit B that A was 
measured to be a 1 or a 0, but no longer have the which way 
information in B.

8. Invert the Hadamard gate

Re: Why are laws of physics stable?

2021-07-07 Thread Jason Resch

On Wed, Jul 7, 2021, 12:14 AM 'Brent Meeker' via Everything List <
everything-list@googlegroups.com> wrote:

>
> On 7/6/2021 6:50 PM, Jason Resch wrote:
>
>
>
> On Tue, Jul 6, 2021, 9:39 PM Bruce Kellett  wrote:
>
>> On Wed, Jul 7, 2021 at 11:29 AM Jason Resch  wrote:
>>
>>> On Tue, Jul 6, 2021, 4:07 PM 'Brent Meeker' via Everything List <
>>> everything-list@googlegroups.com> wrote:
>>>
 On 7/6/2021 10:34 AM, Jason Resch wrote:

 On Tue, Jul 6, 2021 at 12:27 PM 'Brent Meeker' via Everything List <
 everything-list@googlegroups.com> wrote:

> And you're never going to find a being that behaves intelligently
> based on information that can be quantum erased.
>
 You need only a quantum computer with enough qubits.

 Can you prove that?  How does this quantum intelligence ever arrive at
 a definite decision?

>>>
>>> Prove? No. But I think I can justify it:
>>>
>>> 1. Quantum computers are Turing equivalent, they can compute anything a
>>> classical computer can.
>>>
>>> 2. Human brains are believed to operate according to physical laws, all
>>> known of which are computable.
>>>
>>> 3. Humans are conscious.
>>>
>>> 4. By any of: Chalmers's principle of "Organizational invariance", or
>>> "multiple realizability", or the "Generalized Anti-Zombie Principle", or
>>> the "computational theory of mind", a functionally equivalent computation
>>> to that of a conscious human brain will be equivalently conscious to that
>>> brain.
>>>
>>> 5. Quantum computers are reversible.
>>>
>>> By 1 & 2, a quantum computer can simulate a human brain. By 3 & 4, such
>>> an emulation will be conscious. By 5 any computation performed by a quantum
>>> computer can be quantum erased by reversing the circuit back to its
>>> starting state.
>>>
>>> It reaches a definite decision by virtue of completing its processing
>>> before ultimately being reversed. This prevents an outside observer from
>>> learning the decision, but it's made nonetheless during the course of the
>>> processing.
>>>
>>
>> How do you know that it has reached a definite decision? Without having
>> it print out some irreversible record? If it prints out a
>> (pseudo-)classical record, the initial state is not recoverable.
>>
>> Bruce
>>
>
> By either:
>
> 1. Analyzing the circuit
>
> But the question is whether such a circuit is possible.
>
Do you disagree with any of the five premises I defined above? If not do
you see a flaw in my reasoning or conclusions? If not, then why shouldn't
such a circuit be possible?


2. Having the circuit do something useful and verifiable (as in my
> factoring example)
>
> How would you know that had a causal connection to the quantum erasable
> knowledge?
>
This is why I had the information pass through the "AI function" before
being used in Shor's algorithm. That way there was a causal connection with
the result that would be communicated to the outside world.


3. Having the circuit output that it did observe a definite value but
> without reporting which value it observed (as in Deutsch's original example)
>
> Again, how do you know such a circuit is possible?   Most quantum
> computations only produce probable answers in a decohered readout.
>
Ignoring the AI aspect this is a simple and non probabilistic circuit:

1. Initialize qubit A to 0
2. Initialize qubit B to 0
3. Put qubit A into superposition of (0 and 1) via Hadamard gate
4. Apply Controlled NOT gate to (A, B) using A as the control bit to
read/copy the bit value of A to the state of B. (B now has a definite value
of 1 OR it has a definite value of 0)
5. Apply Pauli-X (NOT gate) to (B) to flip the bit value of B (it is now
opposite of A).
6. Apply Controlled NOT gate to (A, B), to have the effect of computing (A
XOR B) and storing the result in B. If A was measured in step 4 as 0, B
will now be 1. Otherwise, if A was measured in step 4 to be 1, B will now
be 1. We now have evidence in qubit B that A was measured to be a 1 or a 0,
but no longer have the which way information in B.
8. Invert the Hadamard gate applied to A to restore it to 0.
9. Read the qubits, while initialized to A = 0, B = 0, you will now find A
= 0, B = 1.


Jason

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Re: Why are laws of physics stable?

2021-07-07 Thread smitra


On 07-07-2021 02:45, 'Brent Meeker' via Everything List wrote:

On 7/6/2021 12:21 AM, smitra wrote:

On 05-07-2021 12:18, Bruce Kellett wrote:

On Mon, Jul 5, 2021 at 7:39 PM smitra  wrote:


On 05-07-2021 09:00, Bruce Kellett wrote:

On Mon, Jul 5, 2021 at 2:23 PM smitra  wrote:

I don't think this is actually done in the experiment. What is
observed is the presence or absence of the interference pattern on

the

screen where the balls hit. The photons are not detected. But if,

in

principle, they are of suitable wavelength to resolve the slit
difference, then the interference pattern vanishes. The experiment

is

convincing in that they start wil cold buckyballs which show a

clear

interference pattern. They then gradually heat the balls so that

the

typical wavelength of the photons decreases. This gradually washes

out

the interference pattern. (Because at lower temperatures, the
wavelength distribution of the IR photons is such that a few of

them

have shorter wavelengths.) As the temperature is increased so that
most IR photons have short enough wavelengths, the interference
pattern disappears completely. The paper by Hornberger et al. is

at

arXiv:quant-ph/0412003v2


This is then what I said previously, what you denied, i.e. that you
are
only considering part of the system which is defined by the reduced
density matrix. The complete system of buckyball plus photons will
show
interference, even if the wavelength is small enough to resolve the
slits provided you perform the right sort of measurement on the
balls
and photons.


That is false.


This is easy to see. Denote the buckyball state of a buckball moving 
through the left slit by |L> and moving through the right slit by |R>. 
Suppose that a photon is emitted by the by the buckyballs such that 
the ball moving through the left slit emits a photon in a state |PL> 
that will be orthogonal to the state |PR> of the photon emitted by the 
ball moving through the right slit . The state of the system after the 
ball passes the slits is then:


|psi> = 1/sqrt(2) [|L>|PL> + |R>|PR>]

This state then evolves under unitary time evolution, we can write the 
state just before the ball hits the screen as:


|psi_s> = 1/sqrt(2) [|L_s>|PL_s> + |R_s>|PR_s>]

There is then no interference patter on the screen for the buckyballs 
because |PL_s> and |PR_s> are orthogonal,


In the Bucky Ball experiment there's no interference pattern when the
photons have long wave length. So it's not just a question of the
states being orthogonal.



If there are many photons, then you have large number of inner products 
between the unobserved photon states and the interference pattern then 
gets suppressed as well.




the unitary time evolution preserves the orthogonality of the initial 
states. The probability to observe a buckyball on position x on the 
screen is:


P(x) = ^2 = 1/2 [||^2 + ||^2] + Re[ 
* ]


And the last interference term is zero because  = 0

But if we also observe the photon on another screen and keep the joint 
count for buckyballs landing on spot x on the buckyball screen and for 
photons landing on spot y on the photon screen as a function of x and 
y, then we do have an interference pattern as a function of x for 
fixed y. If we de note by U the unitary time evolution for the photons 
until they hit their screen, and put |PL_t> =U|PL_s> and |PR_t> = 
U|PR_s>, then the probability distribution is:


P(x,y) = ||^2 = 1/2 [||^2||^2 + 
||^2||^2] +Re[ * *]


The interference term Re[ * *] does not 
vanish as it involves evaluating the components of  the buckyball and 
photon states in the position basis and so there is no inner product 
involved anymore. For fixed y the quantity * will have 
some value that will be nonzero in general, so if we keep y fixed then 
there will be an interference term.


The position operators are projections and include decoherence. I
don't think having fixed y will recover the interference pattern, but
I'm not clear on what y is measuring?  Are there just two spots on the
y-screen corresponding to L and R slits?


Y is measuring the position at which the photon is landing on the screen 
for the photon. When you measure y and x then there is an interference 
in the joint probability distribution. For every fixed y, the function 
P(x,y) will show interference as a function of x. You can also say that 
measuring y destroys the which way information present in the photon 
states.








So, we can conclude that invoking escaping IR photons does not male 
any sense in this discussion because all it does is it scrambles the 
interference pattern to make it invisible in a way that allows it to 
be recovered in principle using measurements on those IR photons. You 
can, of course, erase the interference patter by measuring the 
observable for the photons that has |PR> and |PL> as its eigenstates. 
But even in that case the information will still be there in the state 
of all the atoms of the measurement apparatus for the

Re: Why are laws of physics stable?

2021-07-07 Thread smitra

On 06-07-2021 22:29, 'Brent Meeker' via Everything List wrote:

On 7/6/2021 12:49 PM, smitra wrote:

On 06-07-2021 19:34, Jason Resch wrote:

On Tue, Jul 6, 2021 at 12:27 PM 'Brent Meeker' via Everything List
wrote:

And you're never going to find a being that behaves intelligently
based on information that can be quantum erased.
You need only a quantum computer with enough qubits.

Jason

Indeed, the critics have to show how the laws of physics imply that
decoherence cannot be limited to the extent necessary to run a good
enough quantum computer simulation of an entire brain for this to
work. And one has lots of elbowroom available for the thought
experiment. Practical issue that would make this unfeasible for us to
do play no role at all, but real physical limits would be valid
objections. The amount of available resources that can be used
physically is at least a large fraction of all the materials that are
present in our galaxy. One can build Dyson spheres around a far
fraction of all stars in the galaxy, the available time is at least of
the order of tens of billions of years. The simulation does not have
to run in real time, each simulated second can take a billion years,
which may be necessary to perform enough quantum error correction to
make this work.

If it can be shown that under more generous conditions this is not
feasible, so large scale quantum computing is not going to work even
with most of the resources in the observable universe, and that a
large scale computation needed for the thought experiment cannot be
finished before the end of the universe, then the critics have a
point. But even then it's only a hint of a problem, because the
objection would only be consistent with the unproven hypothesis that
unitary time evolution breaks down when a large enough number of
degrees of freedom get entangled with a quantum system.

Saibal

Why are you worrying about enormous quantum computers? A quantum
computer should have much more computational power than a classical
computer and we already know of an intelligent classical computer fits
in a little more than a liter. The problem isn't computational power,
it's reaching definite answer. Quantum computers in general provide a
readout by decoherence, and then it's no longer erasable.

Brent

There can never be a definite answer as QM is unitary and decoherence is
never complete. If you assume that the real world is fundamentally
different from a virtual world simulated by a quantum computer, no
matter how large that quantum simulation, then you are assuming that the
real world violates QM in an essential way.

Saibal

Re: Why are laws of physics stable?

2021-07-07 Thread smitra

On 07-07-2021 01:17, Bruce Kellett wrote:

On Tue, Jul 6, 2021 at 11:41 PM smitra wrote:

On 06-07-2021 14:10, Bruce Kellett wrote:

It is a shame that your fancy analysis is contradicted by the

actual

experimental results. I leave it as an exercise for you to

determine

where your mistake is. But I suggest that you actually reads the

paper

quant-ph/0412003.

The analysis is 100% rigorous, based on standard QM. The results of
the
paper quant-ph/0412003 are irrelevant as they don't measure the
joint
interference pattern.

It is only your confused notion that even considers such a thing to be
relevant. Hornberger et al. give a complete and detailed quantum
analysis of their experiment and show that their results are in
complete accordance with quantum theory. The thing is that you have to
take their experiment results, particularly as shown in Fig 8 of the
paper, seriously. They show that the interference fringes are
gradually washed out as the temperature of the balls increases. I
quote their abstract:

"We study C_70 fullerene matter waves in a Talbot-Lau interferometer
as a function of their temperature. While the ideal fringe visibility
is observed at moderate molecular temperatures, we find a gradual
degradation of the interference contrast if the molecules are heated
before entering the interferometer. A method is developed to assess
the distribution of the micro-canonical temperature of the molecules
in free flight. This way the heating-dependent reduction of the
interference contrast can be compared with the predictions of quantum
theory. We find that the observed loss of coherence agrees
quantitatively with the expected decoherence rate due to the thermal
radiation emitted by the hot molecules."

None of this contradicts what I wrote.

In that case the interference pattern of the balls
will vanish due to the photon states being orthogonal, as I've shown

above.

The argument against the existence of parallel worlds by invoking
decoherence that makes superposition hard to detect for complex
systems is thus analogous to the defense of creationists when

they invoke a

God of ever smaller gaps of things that have not yet been fully

explained.

My dear, you really have lost the plot, haven't you Saibal?

I'm sticking to QM, your position depends on some unproven effect
that
would make pure states evolve into mixed states.

Your idea of QM is sadly flawed. The real professional quantum
analysis given in the quoted paper shows how the observed effects are
completely consistent with quantum mechanics. The emission of thermal
radiation by the heated balls leads to a clear and evident loss of
coherence. Your pseudo-analysis has nothing to do with either quantum
mechanics or the actual set-up of this buckyball experiment.

As I said, what I wrote is 100% consistent with their results. The fact
that you can't see this implies that you have forgotten basic QM and are
unqualified to continue this discussion. I remember from a while back on
this list that you claimed that in two slit experiment, if the
wavefunctions of the particle moving through the left slit is orthogonal
to that of the right slit, then there would be no interference. This is
obviously wrong, and there was a long discussion where you kept on
repeating your flawed point.

Saibal

Bruce

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Re: Why are laws of physics stable?

2021-07-06 Thread 'Brent Meeker' via Everything List

On 7/6/2021 6:50 PM, Jason Resch wrote:

On Tue, Jul 6, 2021, 9:39 PM Bruce Kellett > wrote:

On Wed, Jul 7, 2021 at 11:29 AM Jason Resch mailto:jasonre...@gmail.com>> wrote:

On Tue, Jul 6, 2021, 4:07 PM 'Brent Meeker' via Everything
List mailto:everything-list@googlegroups.com>> wrote:

On 7/6/2021 10:34 AM, Jason Resch wrote:

On Tue, Jul 6, 2021 at 12:27 PM 'Brent Meeker' via
Everything List mailto:everything-list@googlegroups.com>> wrote:

And you're never going to find a being that behaves
intelligently based on information that can be
quantum erased.

You need only a quantum computer with enough qubits.

Can you prove that? How does this quantum intelligence
ever arrive at a definite decision?

Prove? No. But I think I can justify it:

1. Quantum computers are Turing equivalent, they can compute
anything a classical computer can.

2. Human brains are believed to operate according to physical
laws, all known of which are computable.

3. Humans are conscious.

4. By any of: Chalmers's principle of "Organizational
invariance", or "multiple realizability", or the "Generalized
Anti-Zombie Principle", or the "computational theory of mind",
a functionally equivalent computation to that of a conscious
human brain will be equivalently conscious to that brain.

5. Quantum computers are reversible.

By 1 & 2, a quantum computer can simulate a human brain. By 3
& 4, such an emulation will be conscious. By 5 any computation
performed by a quantum computer can be quantum erased by
reversing the circuit back to its starting state.

It reaches a definite decision by virtue of completing its
processing before ultimately being reversed. This prevents an
outside observer from learning the decision, but it's made
nonetheless during the course of the processing.

How do you know that it has reached a definite decision? Without
having it print out some irreversible record? If it prints out a
(pseudo-)classical record, the initial state is not recoverable.

Bruce

By either:

1. Analyzing the circuit

But the question is whether such a circuit is possible.

2. Having the circuit do something useful and verifiable (as in my
factoring example)

How would you know that had a causal connection to the quantum erasable
knowledge?

3. Having the circuit output that it did observe a definite value but
without reporting which value it observed (as in Deutsch's original
example)

Again, how do you know such a circuit is possible? Most quantum
computations only produce probable answers in a decohered readout.

Brent

Jason

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Re: Why are laws of physics stable?

On Tue, Jul 6, 2021, 11:06 PM Bruce Kellett  wrote:

> On Wed, Jul 7, 2021 at 11:46 AM Jason Resch  wrote:
>
>> On Tue, Jul 6, 2021, 9:22 PM Bruce Kellett  wrote:
>>
>>> On Wed, Jul 7, 2021 at 11:13 AM Jason Resch 
>>> wrote:
>>>
 On Tue, Jul 6, 2021, 2:03 PM 'Brent Meeker' via Everything List <
 everything-list@googlegroups.com> wrote:

> Then I guess I don't understand this part:
>
> *Run it together with Shors algorithm and have "each AI" read a
> definite random number from 0 to 2^n where n is the number of qubits 
> needed
> to represent the semiprime being factored. Then have the AI copy that
> number to another register to prove it went through the AI's mind.*
>
> What does it mean to "read a definite random number"
>
 F(x) is a quantum algorithm (a combination AI + Shor's algorithm) which
 takes an input x where x is a set of N qubits, with each qubit initialized
 to a superposition of 1 and 0.

 Since the qubits are in a superposition representing 2^N states, the
 quantum algorithm likewise becomes a superposition of 2^N uniquely
 processed values. Each one can be viewed as a unique evaluation of F(i)
 where i is each of the possible N-bit bit strings.

 Since F() includes a conscious AI evaluating the input value, and since
 it exists in a superposition, then the evaluation on a quantum computer
 corresponds to 2^N independent conscious states.

 and what does that have to do with recording which slit a photon went
> thru?
>

 It's an alternate example of Deutsch's experiment which shows that
 consciousness doesn't cause collapse, assuming adding a conscious AI to
 Shor's algorithm doesn't somehow break the algorithm. If you can still
 factor numbers with the AI added to the circuit, then consciousness doesn't
 cause collapse, and we can see QM directly leads to many "split" observers.

>>>
>>>
>>> No one now  believes that consciousness has anything to do with
>>> collapse. For example, in fGRW, the collapse is caused by independent
>>> stochastic 'flashes' that have no relevance to consciousness.
>>>
>>
>> Then it says large quantum computers aren't possible. How large dies fGRW
>> say quantum computers can get before they fail? If it gives such a
>> prediction then we can test it. If it doesn't give a prediction it's an
>> empty theory as it's irrefutable and untestable.
>>
>
> First design me your large QC. As far as fGRW is concerned, if the QC is
> defined in terms purely of its qubits, then the limit would probably be
> something on the order of Avogadro's Number of qubits.
>

We should be able to implement a human brain emulation with between 10^15
and 10^18 qubits. A mouse brain would be ~10,000 times less than that.

>> In Bohm's theory, there never is any collapse because there is never any
>>> mystic 'superposition'.
>>>
>>
>> Bohm admitted privately that his theory was a many-worlds theory. I don't
>> have the reference on hand but can try to find it if you're interested.
>>
>
> Deutsch's calumny of Bohm's theory by saying that it was just many worlds
> in denial shows that Deutsch did not really understand Bohm's theory. The
> point of that theory is that particles have definite positions. And there
> is only one position for each particle -- guided by the pilot wave. Many
> worlds would require separate particles for each component of the guiding
> wave. This Bohm denies.
>

What guides the pilot wave, but unseen swaths of "not really there"
particles, all superposing and interfering with each other and behaving
exactly like "really real particles" in that what blocks or what permits
the transmission of them along their many possible paths is exactly the
same as what blocks or transmits the really real particles. There are
collections of not really there particles as elaborate as whole bodies and
brains, and earths, but we need never worry about them, because they're not
real so they can't be conscious, even though you will find books about
consciousness written by interacting collections of these not real
particles in their not really real universes which happen to function to
interfere with the really real particles in our universe. I hope we
ourselves are not made of those fake particles who only make themselves
known by guiding the pilot wave in the real universe.

>
>> In Penroses gravitational induced collapse, the collapse is due to
>>> changes in the spacetime metric -- again, independent of consciousness.
>>>
>>
>> The same critique I made of fGRW applies here. What's does gravitational
>> induced collapse suggest for an upper limit of qubits?
>>
>
> Avogadro's Number?
>

What's special about this number? Does the theory really bear Avagadro's
Number out as a point of significance? I would think something more
fundamental and natural, like a Planck unit would be involved rather than
the number of carbon atoms it takes to

Re: Why are laws of physics stable?

On Tue, Jul 6, 2021, 10:15 PM Bruce Kellett  wrote:

> On Wed, Jul 7, 2021 at 11:50 AM Jason Resch  wrote:
>
>> On Tue, Jul 6, 2021, 9:39 PM Bruce Kellett  wrote:
>>
>>> On Wed, Jul 7, 2021 at 11:29 AM Jason Resch 
>>> wrote:
>>>
 On Tue, Jul 6, 2021, 4:07 PM 'Brent Meeker' via Everything List <
 everything-list@googlegroups.com> wrote:

> On 7/6/2021 10:34 AM, Jason Resch wrote:
>
> On Tue, Jul 6, 2021 at 12:27 PM 'Brent Meeker' via Everything List <
> everything-list@googlegroups.com> wrote:
>
>> And you're never going to find a being that behaves intelligently
>> based on information that can be quantum erased.
>>
> You need only a quantum computer with enough qubits.
>
> Can you prove that?  How does this quantum intelligence ever arrive at
> a definite decision?
>

 Prove? No. But I think I can justify it:

 1. Quantum computers are Turing equivalent, they can compute anything a
 classical computer can.

 2. Human brains are believed to operate according to physical laws, all
 known of which are computable.

 3. Humans are conscious.

 4. By any of: Chalmers's principle of "Organizational invariance", or
 "multiple realizability", or the "Generalized Anti-Zombie Principle", or
 the "computational theory of mind", a functionally equivalent computation
 to that of a conscious human brain will be equivalently conscious to that
 brain.

 5. Quantum computers are reversible.

 By 1 & 2, a quantum computer can simulate a human brain. By 3 & 4, such
 an emulation will be conscious. By 5 any computation performed by a quantum
 computer can be quantum erased by reversing the circuit back to its
 starting state.

 It reaches a definite decision by virtue of completing its processing
 before ultimately being reversed. This prevents an outside observer from
 learning the decision, but it's made nonetheless during the course of the
 processing.

>>>
>>> How do you know that it has reached a definite decision? Without having
>>> it print out some irreversible record? If it prints out a
>>> (pseudo-)classical record, the initial state is not recoverable.
>>>
>>> Bruce
>>>
>>
>> By either:
>>
>> 1. Analyzing the circuit
>> 2. Having the circuit do something useful and verifiable (as in my
>> factoring example)
>> 3. Having the circuit output that it did observe a definite value but
>> without reporting which value it observed (as in Deutsch's original example)
>>
>
>
> All of these involve decoherence.
>

#1. Is just looking at the program and verifying if the quantum computer
runs it, it will reach a particular state in the course of it's operation,
in the same way we might looks at the experimental setup of a quantum
eraser setup and conclude when the photon is in this position in the
apparatus that it will be in a particular state, even if that state is
ultimately erased from our view later.

You have to show that you can decohere part of your QC without decohering
> the rest. I wish you luck!
>

#2. This is done by Shor's algorithm, which erases parts of the computation
so other parts can successfully interfere to give the desired answer.
Shor's algorithm has been implemented and run.

#3. Would be another example of #2, reversing (and erasing) a prior
measurement of a qubit put in a superposition after storing a bit
indication that the qubit was a 0 or 1 (but not saying which). I believe
Deutsch describes this experiment in his seminal paper that defined the
universal quantum computer in 1985.

Jason

> Bruce
>
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> .
>

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Re: Why are laws of physics stable?

On Wed, Jul 7, 2021 at 11:46 AM Jason Resch  wrote:

> On Tue, Jul 6, 2021, 9:22 PM Bruce Kellett  wrote:
>
>> On Wed, Jul 7, 2021 at 11:13 AM Jason Resch  wrote:
>>
>>> On Tue, Jul 6, 2021, 2:03 PM 'Brent Meeker' via Everything List <
>>> everything-list@googlegroups.com> wrote:
>>>
 Then I guess I don't understand this part:

 *Run it together with Shors algorithm and have "each AI" read a
 definite random number from 0 to 2^n where n is the number of qubits needed
 to represent the semiprime being factored. Then have the AI copy that
 number to another register to prove it went through the AI's mind.*

 What does it mean to "read a definite random number"

>>> F(x) is a quantum algorithm (a combination AI + Shor's algorithm) which
>>> takes an input x where x is a set of N qubits, with each qubit initialized
>>> to a superposition of 1 and 0.
>>>
>>> Since the qubits are in a superposition representing 2^N states, the
>>> quantum algorithm likewise becomes a superposition of 2^N uniquely
>>> processed values. Each one can be viewed as a unique evaluation of F(i)
>>> where i is each of the possible N-bit bit strings.
>>>
>>> Since F() includes a conscious AI evaluating the input value, and since
>>> it exists in a superposition, then the evaluation on a quantum computer
>>> corresponds to 2^N independent conscious states.
>>>
>>>
>>> and what does that have to do with recording which slit a photon went
 thru?

>>>
>>> It's an alternate example of Deutsch's experiment which shows that
>>> consciousness doesn't cause collapse, assuming adding a conscious AI to
>>> Shor's algorithm doesn't somehow break the algorithm. If you can still
>>> factor numbers with the AI added to the circuit, then consciousness doesn't
>>> cause collapse, and we can see QM directly leads to many "split" observers.
>>>
>>
>>
>> No one now  believes that consciousness has anything to do with collapse.
>> For example, in fGRW, the collapse is caused by independent stochastic
>> 'flashes' that have no relevance to consciousness.
>>
>
> Then it says large quantum computers aren't possible. How large dies fGRW
> say quantum computers can get before they fail? If it gives such a
> prediction then we can test it. If it doesn't give a prediction it's an
> empty theory as it's irrefutable and untestable.
>

First design me your large QC. As far as fGRW is concerned, if the QC is
defined in terms purely of its qubits, then the limit would probably be
something on the order of Avogadro's Number of qubits.

>
> In Bohm's theory, there never is any collapse because there is never any
>> mystic 'superposition'.
>>
>
> Bohm admitted privately that his theory was a many-worlds theory. I don't
> have the reference on hand but can try to find it if you're interested.
>

Deutsch's calumny of Bohm's theory by saying that it was just many worlds
in denial shows that Deutsch did not really understand Bohm's theory. The
point of that theory is that particles have definite positions. And there
is only one position for each particle -- guided by the pilot wave. Many
worlds would require separate particles for each component of the guiding
wave. This Bohm denies.


> In Penroses gravitational induced collapse, the collapse is due to changes
>> in the spacetime metric -- again, independent of consciousness.
>>
>
> The same critique I made of fGRW applies here. What's does gravitational
> induced collapse suggest for an upper limit of qubits?
>

Avogadro's Number?

 So Deutsch's thought experiment is about nothing at all, and proves
>> nothing at all.
>>
>
> Deutsch's aim was to show that collapse vs. no collapse was in principle
> testable. All you write above confirms this as these spontaneous collapse
> theories make different predictions which are testable, so they're not just
> different interpretations, but different theories.
>

I agree that they are different theories. And that all are, in principle,
testable. But then, as I said to Saibal, it is unlikely that unitary
evolution according to the Schrodinger equation is the final theory either.
Deutsch's thought experiment proves nothing about MWI, one way or the other.

Bruce

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Re: Why are laws of physics stable?

On Wed, Jul 7, 2021 at 11:50 AM Jason Resch  wrote:

> On Tue, Jul 6, 2021, 9:39 PM Bruce Kellett  wrote:
>
>> On Wed, Jul 7, 2021 at 11:29 AM Jason Resch  wrote:
>>
>>> On Tue, Jul 6, 2021, 4:07 PM 'Brent Meeker' via Everything List <
>>> everything-list@googlegroups.com> wrote:
>>>
 On 7/6/2021 10:34 AM, Jason Resch wrote:

 On Tue, Jul 6, 2021 at 12:27 PM 'Brent Meeker' via Everything List <
 everything-list@googlegroups.com> wrote:

> And you're never going to find a being that behaves intelligently
> based on information that can be quantum erased.
>
 You need only a quantum computer with enough qubits.

 Can you prove that?  How does this quantum intelligence ever arrive at
 a definite decision?

>>>
>>> Prove? No. But I think I can justify it:
>>>
>>> 1. Quantum computers are Turing equivalent, they can compute anything a
>>> classical computer can.
>>>
>>> 2. Human brains are believed to operate according to physical laws, all
>>> known of which are computable.
>>>
>>> 3. Humans are conscious.
>>>
>>> 4. By any of: Chalmers's principle of "Organizational invariance", or
>>> "multiple realizability", or the "Generalized Anti-Zombie Principle", or
>>> the "computational theory of mind", a functionally equivalent computation
>>> to that of a conscious human brain will be equivalently conscious to that
>>> brain.
>>>
>>> 5. Quantum computers are reversible.
>>>
>>> By 1 & 2, a quantum computer can simulate a human brain. By 3 & 4, such
>>> an emulation will be conscious. By 5 any computation performed by a quantum
>>> computer can be quantum erased by reversing the circuit back to its
>>> starting state.
>>>
>>> It reaches a definite decision by virtue of completing its processing
>>> before ultimately being reversed. This prevents an outside observer from
>>> learning the decision, but it's made nonetheless during the course of the
>>> processing.
>>>
>>
>> How do you know that it has reached a definite decision? Without having
>> it print out some irreversible record? If it prints out a
>> (pseudo-)classical record, the initial state is not recoverable.
>>
>> Bruce
>>
>
> By either:
>
> 1. Analyzing the circuit
> 2. Having the circuit do something useful and verifiable (as in my
> factoring example)
> 3. Having the circuit output that it did observe a definite value but
> without reporting which value it observed (as in Deutsch's original example)
>


All of these involve decoherence. You have to show that you can decohere
part of your QC without decohering the rest. I wish you luck!

Bruce

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Re: Why are laws of physics stable?

On Tue, Jul 6, 2021, 9:39 PM Bruce Kellett  wrote:

> On Wed, Jul 7, 2021 at 11:29 AM Jason Resch  wrote:
>
>> On Tue, Jul 6, 2021, 4:07 PM 'Brent Meeker' via Everything List <
>> everything-list@googlegroups.com> wrote:
>>
>>> On 7/6/2021 10:34 AM, Jason Resch wrote:
>>>
>>> On Tue, Jul 6, 2021 at 12:27 PM 'Brent Meeker' via Everything List <
>>> everything-list@googlegroups.com> wrote:
>>>
 And you're never going to find a being that behaves intelligently based
 on information that can be quantum erased.

>>> You need only a quantum computer with enough qubits.
>>>
>>> Can you prove that?  How does this quantum intelligence ever arrive at a
>>> definite decision?
>>>
>>
>> Prove? No. But I think I can justify it:
>>
>> 1. Quantum computers are Turing equivalent, they can compute anything a
>> classical computer can.
>>
>> 2. Human brains are believed to operate according to physical laws, all
>> known of which are computable.
>>
>> 3. Humans are conscious.
>>
>> 4. By any of: Chalmers's principle of "Organizational invariance", or
>> "multiple realizability", or the "Generalized Anti-Zombie Principle", or
>> the "computational theory of mind", a functionally equivalent computation
>> to that of a conscious human brain will be equivalently conscious to that
>> brain.
>>
>> 5. Quantum computers are reversible.
>>
>> By 1 & 2, a quantum computer can simulate a human brain. By 3 & 4, such
>> an emulation will be conscious. By 5 any computation performed by a quantum
>> computer can be quantum erased by reversing the circuit back to its
>> starting state.
>>
>> It reaches a definite decision by virtue of completing its processing
>> before ultimately being reversed. This prevents an outside observer from
>> learning the decision, but it's made nonetheless during the course of the
>> processing.
>>
>
> How do you know that it has reached a definite decision? Without having it
> print out some irreversible record? If it prints out a (pseudo-)classical
> record, the initial state is not recoverable.
>
> Bruce
>

By either:

1. Analyzing the circuit
2. Having the circuit do something useful and verifiable (as in my
factoring example)
3. Having the circuit output that it did observe a definite value but
without reporting which value it observed (as in Deutsch's original example)

Jason



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Re: Why are laws of physics stable?

On Tue, Jul 6, 2021, 9:22 PM Bruce Kellett  wrote:

> On Wed, Jul 7, 2021 at 11:13 AM Jason Resch  wrote:
>
>> On Tue, Jul 6, 2021, 2:03 PM 'Brent Meeker' via Everything List <
>> everything-list@googlegroups.com> wrote:
>>
>>> Then I guess I don't understand this part:
>>>
>>> *Run it together with Shors algorithm and have "each AI" read a definite
>>> random number from 0 to 2^n where n is the number of qubits needed to
>>> represent the semiprime being factored. Then have the AI copy that number
>>> to another register to prove it went through the AI's mind.*
>>>
>>> What does it mean to "read a definite random number"
>>>
>> F(x) is a quantum algorithm (a combination AI + Shor's algorithm) which
>> takes an input x where x is a set of N qubits, with each qubit initialized
>> to a superposition of 1 and 0.
>>
>> Since the qubits are in a superposition representing 2^N states, the
>> quantum algorithm likewise becomes a superposition of 2^N uniquely
>> processed values. Each one can be viewed as a unique evaluation of F(i)
>> where i is each of the possible N-bit bit strings.
>>
>> Since F() includes a conscious AI evaluating the input value, and since
>> it exists in a superposition, then the evaluation on a quantum computer
>> corresponds to 2^N independent conscious states.
>>
>>
>> and what does that have to do with recording which slit a photon went
>>> thru?
>>>
>>
>> It's an alternate example of Deutsch's experiment which shows that
>> consciousness doesn't cause collapse, assuming adding a conscious AI to
>> Shor's algorithm doesn't somehow break the algorithm. If you can still
>> factor numbers with the AI added to the circuit, then consciousness doesn't
>> cause collapse, and we can see QM directly leads to many "split" observers.
>>
>
>
> No one now  believes that consciousness has anything to do with collapse.
> For example, in fGRW, the collapse is caused by independent stochastic
> 'flashes' that have no relevance to consciousness.
>

Then it says large quantum computers aren't possible. How large dies fGRW
say quantum computers can get before they fail? If it gives such a
prediction then we can test it. If it doesn't give a prediction it's an
empty theory as it's irrefutable and untestable.

In Bohm's theory, there never is any collapse because there is never any
> mystic 'superposition'.
>

Bohm admitted privately that his theory was a many-worlds theory. I don't
have the reference on hand but can try to find it if you're interested.

In Penroses gravitational induced collapse, the collapse is due to changes
> in the spacetime metric -- again, independent of consciousness.
>

The same critique I made of fGRW applies here. What's does gravitational
induced collapse suggest for an upper limit of qubits?


>  So Deutsch's thought experiment is about nothing at all, and proves
> nothing at all.
>

Deutsch's aim was to show that collapse vs. no collapse was in principle
testable. All you write above confirms this as these spontaneous collapse
theories make different predictions which are testable, so they're not just
different interpretations, but different theories.

Jason


> Bruce
>
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Re: Why are laws of physics stable?

On Wed, Jul 7, 2021 at 11:29 AM Jason Resch  wrote:

> On Tue, Jul 6, 2021, 4:07 PM 'Brent Meeker' via Everything List <
> everything-list@googlegroups.com> wrote:
>
>> On 7/6/2021 10:34 AM, Jason Resch wrote:
>>
>> On Tue, Jul 6, 2021 at 12:27 PM 'Brent Meeker' via Everything List <
>> everything-list@googlegroups.com> wrote:
>>
>>> And you're never going to find a being that behaves intelligently based
>>> on information that can be quantum erased.
>>>
>> You need only a quantum computer with enough qubits.
>>
>> Can you prove that?  How does this quantum intelligence ever arrive at a
>> definite decision?
>>
>
> Prove? No. But I think I can justify it:
>
> 1. Quantum computers are Turing equivalent, they can compute anything a
> classical computer can.
>
> 2. Human brains are believed to operate according to physical laws, all
> known of which are computable.
>
> 3. Humans are conscious.
>
> 4. By any of: Chalmers's principle of "Organizational invariance", or
> "multiple realizability", or the "Generalized Anti-Zombie Principle", or
> the "computational theory of mind", a functionally equivalent computation
> to that of a conscious human brain will be equivalently conscious to that
> brain.
>
> 5. Quantum computers are reversible.
>
> By 1 & 2, a quantum computer can simulate a human brain. By 3 & 4, such an
> emulation will be conscious. By 5 any computation performed by a quantum
> computer can be quantum erased by reversing the circuit back to its
> starting state.
>
> It reaches a definite decision by virtue of completing its processing
> before ultimately being reversed. This prevents an outside observer from
> learning the decision, but it's made nonetheless during the course of the
> processing.
>

How do you know that it has reached a definite decision? Without having it
print out some irreversible record? If it prints out a (pseudo-)classical
record, the initial state is not recoverable.

Bruce

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Re: Why are laws of physics stable?

On Tue, Jul 6, 2021, 4:07 PM 'Brent Meeker' via Everything List <
everything-list@googlegroups.com> wrote:

>
> On 7/6/2021 10:34 AM, Jason Resch wrote:
>
>
>
> On Tue, Jul 6, 2021 at 12:27 PM 'Brent Meeker' via Everything List <
> everything-list@googlegroups.com> wrote:
>
>> And you're never going to find a being that behaves intelligently based
>> on information that can be quantum erased.
>>
> You need only a quantum computer with enough qubits.
>
> Can you prove that?  How does this quantum intelligence ever arrive at a
> definite decision?
>

Prove? No. But I think I can justify it:

1. Quantum computers are Turing equivalent, they can compute anything a
classical computer can.

2. Human brains are believed to operate according to physical laws, all
known of which are computable.

3. Humans are conscious.

4. By any of: Chalmers's principle of "Organizational invariance", or
"multiple realizability", or the "Generalized Anti-Zombie Principle", or
the "computational theory of mind", a functionally equivalent computation
to that of a conscious human brain will be equivalently conscious to that
brain.

5. Quantum computers are reversible.

By 1 & 2, a quantum computer can simulate a human brain. By 3 & 4, such an
emulation will be conscious. By 5 any computation performed by a quantum
computer can be quantum erased by reversing the circuit back to its
starting state.

It reaches a definite decision by virtue of completing its processing
before ultimately being reversed. This prevents an outside observer from
learning the decision, but it's made nonetheless during the course of the
processing.

Jason

Brent
>
>
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Re: Why are laws of physics stable?

On Wed, Jul 7, 2021 at 11:13 AM Jason Resch  wrote:

> On Tue, Jul 6, 2021, 2:03 PM 'Brent Meeker' via Everything List <
> everything-list@googlegroups.com> wrote:
>
>> Then I guess I don't understand this part:
>>
>> *Run it together with Shors algorithm and have "each AI" read a definite
>> random number from 0 to 2^n where n is the number of qubits needed to
>> represent the semiprime being factored. Then have the AI copy that number
>> to another register to prove it went through the AI's mind.*
>>
>> What does it mean to "read a definite random number"
>>
> F(x) is a quantum algorithm (a combination AI + Shor's algorithm) which
> takes an input x where x is a set of N qubits, with each qubit initialized
> to a superposition of 1 and 0.
>
> Since the qubits are in a superposition representing 2^N states, the
> quantum algorithm likewise becomes a superposition of 2^N uniquely
> processed values. Each one can be viewed as a unique evaluation of F(i)
> where i is each of the possible N-bit bit strings.
>
> Since F() includes a conscious AI evaluating the input value, and since it
> exists in a superposition, then the evaluation on a quantum computer
> corresponds to 2^N independent conscious states.
>
>
> and what does that have to do with recording which slit a photon went thru?
>>
>
> It's an alternate example of Deutsch's experiment which shows that
> consciousness doesn't cause collapse, assuming adding a conscious AI to
> Shor's algorithm doesn't somehow break the algorithm. If you can still
> factor numbers with the AI added to the circuit, then consciousness doesn't
> cause collapse, and we can see QM directly leads to many "split" observers.
>


No one now  believes that consciousness has anything to do with collapse.
For example, in fGRW, the collapse is caused by independent stochastic
'flashes' that have no relevance to consciousness. In Bohm's theory, there
never is any collapse because there is never any mystic 'superposition'. In
Penroses gravitational induced collapse, the collapse is due to changes in
the spacetime metric -- again, independent of consciousness.

 So Deutsch's thought experiment is about nothing at all, and proves
nothing at all.

Bruce

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Re: Why are laws of physics stable?

On Tue, Jul 6, 2021, 2:03 PM 'Brent Meeker' via Everything List <
everything-list@googlegroups.com> wrote:

>
> On 7/6/2021 6:50 AM, Jason Resch wrote:
>
>
>
> On Mon, Jul 5, 2021, 2:52 PM 'Brent Meeker' via Everything List <
> everything-list@googlegroups.com> wrote:
>
>>
>> On 7/5/2021 5:46 AM, Jason Resch wrote:
>>
>>
>>
>> On Sun, Jul 4, 2021, 8:41 PM 'Brent Meeker' via Everything List <
>> everything-list@googlegroups.com> wrote:
>>
>>>
>>> On 7/4/2021 5:05 PM, Jason Resch wrote:
>>>
>>>
>>>
>>> On Sun, Jul 4, 2021, 3:36 PM 'Brent Meeker' via Everything List <
>>> everything-list@googlegroups.com> wrote:
>>>

 On 7/4/2021 8:01 AM, John Clark wrote:

 On Sun, Jul 4, 2021 at 9:07 AM Lawrence Crowell <
 goldenfieldquaterni...@gmail.com> wrote:

> > *I can imagine this being worked without MWI. The nonlocality of
> the gravitation field and the locality of QFT means that with spacetime
> formed by entanglements of quantum states or fields, that locality and
> nonlocality may be shifted around. Decoherence and the transition of a
> quantum state or entanglement to a decoherent set may be thought of as a
> nonlocal process.*
>

 Maybe the above can be imagined, but it's a whole lot easier imagining
 many worlds. I keep thinking of epicycles in astronomy, one needs to
 go through a lot of strenuous mental gymnastics to avoid the obvious
 conclusion that many worlds exist.

 > *This may be worked so the objective collapse in GRW is such a
> shift. *
>

 I think GRW should be ruled out by Occam's razor, it requires extra
 terms be added to Schrodinger's equation which make it more difficult to
 solve and do not improve its ability to make predictions of observable
 events, in fact it makes the predictions worse because unlike Dirac's 
 Equation
 or Many Worlds it is not compatible with Special Relativity.

  > *There are quantum interpretations that are ψ-epistemic, Copenhagen
> Interpretation, Qubism etc and those that are ψ-ontic such as Many Worlds
> or Bohm interpretations. I think there is no decision procedure that can
> ever tell us which of these sets quantum physics sets within. I would then
> say which ever one of these you work with is a matter of your choice. I
> suspect there is no way we can ever know for sure which of these is
> correct,*
>

 I think I mentioned before that in David Deutsch's book "The Ghost In
 The Atom" he proposed an experimental test that would be very difficult
 , but not impossible, to perform that could decide between Copenhagen
 and Many Worlds; and the reason it's so difficult is not Many Worlds
 fault, the reason is that the conventional view says conscious observers
 obey different laws of physics, Many Worlds says they do not, so to test
 who's right we need a mind that uses quantum properties and
 algorithms.

 An intelligent quantum computer shoots photons at a metal plate one at
 a time that has 2 small slits in it, and then the photons hit a
 photographic plate. Nobody looks at the photographic plate till the very
 end of the experiment. The quantum mind has detectors near each slit so it
 knows which slit the various photons went through. After each photon passes
 the slits, but before they hit the photographic plate, the quantum
 mind signs a document saying that it has observed each and every photon and
 knows which slit each photon went through. It is very important that the
 document does NOT say which slit a photon went through, it only says that
 it went through one slit and only one slit and the mind has knowledge
 of which one. There is a signed document to this effect for every photon it
 shoots.

 Now the mind uses quantum erasure to completely destroy its memory of
 which slit any of the photons went through; the only part remaining in the
 universe is the document which states that each photon went through one and
 only one slit and the mind (at the time) knew which one. Now develop the
 photographic plate and look at it. If you see interference bands then the
 Many World interpretation is correct. If you do not see interference bands
 then there are no worlds but this one and the conventional quantum
 interpretation is correct.

 This works because in the Copenhagen interpretation when the results of
 a measurement enters the consciousness of an observer the wave function
 collapses, in effect all the universes except one disappear without a trace
 so you get no interference. In the Many Worlds model all the other worlds
 will converge back into one universe because information on which slit the
 various photons went through was the only thing that made one universe
 different from another, so when that was erased they became identical again

Re: Why are laws of physics stable?

2021-07-06 Thread 'Brent Meeker' via Everything List




On 7/6/2021 12:21 AM, smitra wrote:

On 05-07-2021 12:18, Bruce Kellett wrote:

On Mon, Jul 5, 2021 at 7:39 PM smitra  wrote:


On 05-07-2021 09:00, Bruce Kellett wrote:

On Mon, Jul 5, 2021 at 2:23 PM smitra  wrote:

I don't think this is actually done in the experiment. What is
observed is the presence or absence of the interference pattern on

the

screen where the balls hit. The photons are not detected. But if,

in

principle, they are of suitable wavelength to resolve the slit
difference, then the interference pattern vanishes. The experiment

is

convincing in that they start wil cold buckyballs which show a

clear

interference pattern. They then gradually heat the balls so that

the

typical wavelength of the photons decreases. This gradually washes

out

the interference pattern. (Because at lower temperatures, the
wavelength distribution of the IR photons is such that a few of

them

have shorter wavelengths.) As the temperature is increased so that
most IR photons have short enough wavelengths, the interference
pattern disappears completely. The paper by Hornberger et al. is

at

arXiv:quant-ph/0412003v2


This is then what I said previously, what you denied, i.e. that you
are
only considering part of the system which is defined by the reduced
density matrix. The complete system of buckyball plus photons will
show
interference, even if the wavelength is small enough to resolve the
slits provided you perform the right sort of measurement on the
balls
and photons.


That is false.


This is easy to see. Denote the buckyball state of a buckball moving 
through the left slit by |L> and moving through the right slit by |R>. 
Suppose that a photon is emitted by the by the buckyballs such that 
the ball moving through the left slit emits a photon in a state |PL> 
that will be orthogonal to the state |PR> of the photon emitted by the 
ball moving through the right slit . The state of the system after the 
ball passes the slits is then:


|psi> = 1/sqrt(2) [|L>|PL> + |R>|PR>]

This state then evolves under unitary time evolution, we can write the 
state just before the ball hits the screen as:


|psi_s> = 1/sqrt(2) [|L_s>|PL_s> + |R_s>|PR_s>]

There is then no interference patter on the screen for the buckyballs 
because |PL_s> and |PR_s> are orthogonal, 


In the Bucky Ball experiment there's no interference pattern when the 
photons have long wave length. So it's not just a question of the states 
being orthogonal.



the unitary time evolution preserves the orthogonality of the initial 
states. The probability to observe a buckyball on position x on the 
screen is:


P(x) = ^2 = 1/2 [||^2 + ||^2] + Re[ 
* ]


And the last interference term is zero because  = 0

But if we also observe the photon on another screen and keep the joint 
count for buckyballs landing on spot x on the buckyball screen and for 
photons landing on spot y on the photon screen as a function of x and 
y, then we do have an interference pattern as a function of x for 
fixed y. If we de note by U the unitary time evolution for the photons 
until they hit their screen, and put |PL_t> =U|PL_s> and |PR_t> = 
U|PR_s>, then the probability distribution is:


P(x,y) = ||^2 = 1/2 [||^2||^2 + 
||^2||^2] +Re[ * *]


The interference term Re[ * *] does not 
vanish as it involves evaluating the components of  the buckyball and 
photon states in the position basis and so there is no inner product 
involved anymore. For fixed y the quantity * will have 
some value that will be nonzero in general, so if we keep y fixed then 
there will be an interference term.


The position operators are projections and include decoherence. I don't 
think having fixed y will recover the interference pattern, but I'm not 
clear on what y is measuring?  Are there just two spots on the y-screen 
corresponding to L and R slits?





So, we can conclude that invoking escaping IR photons does not male 
any sense in this discussion because all it does is it scrambles the 
interference pattern to make it invisible in a way that allows it to 
be recovered in principle using measurements on those IR photons. You 
can, of course, erase the interference patter by measuring the 
observable for the photons that has |PR> and |PL> as its eigenstates. 
But even in that case the information will still be there in the state 
of all the atoms of the measurement apparatus for the photons. But if 
you don't perform any measurement then the information will simply 
continue to exists in the escaping photons.


And per the first sentence of the paragraph the interference will be 
eliminated by the escaping IR photons.  Are you contradicting this in 
the last sentence?  The experiment showed the interference  disappeared 
as the IR photon wave length decreased. It said nothing about them being 
observed or escaping into space.


Brent




So, in general we can conclude by generalizing this to any large 
number of particles that even with what we consider to be permanent

Re: Why are laws of physics stable?

On Wed, Jul 7, 2021 at 6:29 AM 'Brent Meeker' via Everything List <
everything-list@googlegroups.com> wrote:

>
> On 7/6/2021 12:49 PM, smitra wrote:
> > On 06-07-2021 19:34, Jason Resch wrote:
> >> On Tue, Jul 6, 2021 at 12:27 PM 'Brent Meeker' via Everything List
> >>  wrote:
> >>
> >>>
> >>
> >> And you're never going to find a being that behaves intelligently
> >> based on information that can be quantum erased.
> >> You need only a quantum computer with enough qubits.
> >>
> >> Jason
> >
> > Indeed, the critics have to show how the laws of physics imply that
> > decoherence cannot be limited to the extent necessary to run a good
> > enough quantum computer simulation of an entire brain for this to
> > work. And one has lots of elbowroom available for the thought
> > experiment. Practical issue that would make this unfeasible for us to
> > do play no role at all, but real physical limits would be valid
> > objections. The amount of available resources that can be used
> > physically is at least a large fraction of all the materials that are
> > present in our galaxy. One can build Dyson spheres around a far
> > fraction of all stars in the galaxy, the available time is at least of
> > the order of tens of billions of years. The simulation does not have
> > to run in real time, each simulated second can take a billion years,
> > which may be necessary to perform enough quantum error correction to
> > make this work.
> >
> > If it can be shown that under more generous conditions this is not
> > feasible, so large scale quantum computing is not going to work even
> > with most of the resources in the observable universe, and that a
> > large scale computation needed for the thought experiment cannot be
> > finished before the end of the universe, then the critics have a
> > point. But even then it's only a hint of a problem, because the
> > objection would only be consistent with the unproven hypothesis that
> > unitary time evolution breaks down when a large enough number of
> > degrees of freedom get entangled with a quantum system.
> >
> > Saibal
>
> Why are you worrying about enormous quantum computers?  A quantum
> computer should have much more computational power than a classical
> computer and we already know of an intelligent classical computer fits
> in a little more than a liter.  The problem isn't computational power,
> it's reaching definite answer.  Quantum computers in general provide a
> readout by decoherence, and then it's no longer erasable.
>

Exactly right, Brent. It is not resources that are necessary, it is an
explanation of how a consciousness that records a definite result can have
that information quantum erased. If one thinks that it is just a matter of
overwriting some computer registers, then there is no way that you can ever
demonstrate that a definite result was ever obtained. Definite results
require decoherence and the formation of permanent records. Since
thermodynamics assures us that nothing  can ever be totally isolated from
the environment, decoherence is ubiquitous. Then the fact that IR photos
will always irreversibly escape from your set-up ensures that definite
quantum results are always irreversible. As Bohr and others would have
said, a definite result is always a classical object -- subject to the
classical laws of relativity and thermodynamics. A quantum computer, no
matter how many qubits it has, or what resources it utilizes, is never
going to escape from these limits imposed by the laws of physics.

The trouble with many computer scientists is that they live in some
idealized world that has nothing to do with the laws of physics.

Bruce

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Re: Why are laws of physics stable?